text:
\section{Introduction}\label{s:smmain}

\subsection{Overview and main results}\label{s:intro-int}

Motivation and background for this paper may be found in Part I \cite[\S1]{CSI}.
We adopt the notation and setting of Part I \cite[\S1.1]{CSI}.
In particular we consider {\it smooth} manifolds, embeddings and isotopies and recall that:

$\bullet$ $N$ is a closed connected orientable $4$-manifold;

$\bullet$ $E^m(N)$ denotes the set of isotopy classes of embeddings $f \colon N \to S^m$.

In this paper we classify $E^7(N)$ when $H_1(N; {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H})$ it torsion free (up to an indeterminancy in certain cases).
See Theorems \ref{t:s1s3sm}, \ref{t:gensm} and Corollaries \ref{t:actioni}, \ref{c:gen} below.
Our classification is complete when $H_2 = 0$ (see Theorem \ref{t:gensm} and Corollary \ref{c:gen}.b) or when the signature of $N$ is divisible neither by 64 nor by 9 (see Theorem \ref{t:gensm} and Corollary \ref{t:actioni}).
The classification requires finding a complete set of invariants and constructing embeddings realizing particular values of these invariants.
The invariants we use are described in \cite[Lemma 1.3, \S2.2, \S2.3]{CSI} and \S\ref{s:defandplan-eta}.
An overview of the proof of their completeness is given in \cite[\S1.4]{CSI} and in Remark \ref{r:action} below.

The group $E^7(S^4)\cong{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$ acts on $E^7(N)$ by embedded connected sum.
This action was investigated in \cite{Sk10} and determined when $H_1(N;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H})= 0$ in \cite{CS11},
which also classified $E^7(N)$ in this case.
In \cite{CSI} we described the quotient $E_{\# }^7(N):=E^7(N)/E^7(S^4)$ when $H_1(N;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H})=0$.
Thus the main novelty of this paper is the description of this action for $H_1(N;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H})\ne0$.
Cf.~Remark \ref{r:general}.

Denote by $q_{\# }:E^7(N)\to E^7_{\# }(N)$ the quotient map.

In order to state our main result for $N=S^1\times S^3$ consider the following diagram
(where the left triangle is not commutative):
$$
\xymatrix{
{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}\times{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}^2 \ar[r]^{\mathop{\fam0 pr}_2} \ar[d]_{\#\times\tau} & {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}^2 \ar[d]^{\tau_{\# }:=q_{\# }\tau} \ar[dl]^{\tau}\\
E^7(S^1\times S^3)  \ar[r]_{q_\#} & E_{\# }^7(S^1\times S^3).}
$$
The maps $\tau,q_{\# }$ are defined in \cite[\S1.2]{CSI}.
We define
$$
\#\times\tau\quad\text{by}\quad (\#\times\tau)(a,l,b):=a\#\tau(l,b)\quad\text{and}\quad\tau_{\# }:=q_{\# }\tau.
$$

\begin{Theorem} \label{t:s1s3sm}
The map $\#\times\tau\colon {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}\times{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}^2\to E^7(S^1\times S^3)$ is a surjection such that

(a) for different $l,b$ the sets $P_{l,b}:=(\#\times\tau)({\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}\times(l,b))$  either are disjoint or coincide;

$$
(b)\qquad P_{l,b}=P_{l',b'}\quad\Leftrightarrow\quad(\ l=l' \quad\text{and}\quad b\equiv b'\mo2l\ );
$$
$$
(c)\qquad |P_{l,b}|=\begin{cases}12 & l\ne0\\ 2\gcd(b,6) &l=0.\end{cases}
$$
\end{Theorem}

In Theorem \ref{t:s1s3sm} the surjectivity of $\tau$ and (a) and (b) follow from \cite[Theorem 1.1]{CSI}.
The new part of Theorem \ref{t:s1s3sm} is (c); this part follows from
Corollary \ref{c:gen}.b below (because for $l\ne0$ the group $\mathop{\fam0 coker}\overline l$ is finite, so $\mathop{\fam0 div} b=0$).
Cf.~Addendum \ref{c:s1s3}.

\begin{Example}\label{e:com}
There is an embedding $f\colon S^1\times S^3\to S^7$ with $f(N)\subset S^6$ and a pair of non-isotopic embeddings $g_1,g_2\colon S^4\to S^7$  such that $f\#g_1$ and $f\#g_2$ are isotopic.
\end{Example}

This example follows because there is a representative of $\tau(0,1)$ whose image is in $S^6\subset S^7$ \cite[Lemma 2.21]{CSI} and
$|P_{0,1}|=2$ by Theorem \ref{t:s1s3sm}.

Example \ref{e:com} shows necessity of the assumption of simple-connectivity the following result
(which is \cite[The Effectiveness Theorem 1.2]{Sk10}):%
\footnote{It follows that the Effectiveness Theorem 1.2 of the earlier versions of \cite{Sk10} was false.}

{\it If $f\colon N\to S^7$ is an embedding of a spin simply-connected closed 4-manifold $N$, \ $f(N)\subset S^6$ and embeddings $g_1,g_2\colon S^4\to S^7$ are not isotopic, then $f\#g_1$ and $f\#g_2$ are not isotopic.}

Before stating our main result in the general case we state the following corollary of it.

\begin{Corollary}[of Theorem \ref{t:gensm}.c]\label{t:actioni}
(a) If $\varkappa(f)$ is neither divisible by $4$ nor by $3$, then for each embedding $g\colon S^4\to S^7$ the embeddings $f\#g$ and $f$ are isotopic.

(b) If $\varkappa(f)$ is divisible by 4 but neither by 8 nor by 3, then there is a non-trivial embedding $g_1\colon S^4\to S^7$ such that for every  embedding $g\colon S^4\to S^7$ the embedding $f\#g$ is isotopic to either $f$ or $f\#g_1$.
\end{Corollary}

Corollary \ref{t:actioni} follows from Corollary \ref{c:gen}.bc or from Theorem \ref{t:gensm}.c
(because $4{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\gcd(\varkappa(f),24)}=0$ under the assumptions of Corollary \ref{t:actioni}).
The assumption of Corollary \ref{t:actioni}.a is automatically satisfied
when the signature of $N$ is divisible neither by 16 nor by 9.
Cf.~Remark \ref{r:at}.a.

For stating our main result in the general case we use the conventions, notation and definitions of \cite[\S1.2]{CSI}.
Denote
$$
\widehat n:=\gcd(n,24).
$$
{\it If $H_1=0$, then the map $\varkappa_{\# }$ (which is the Bo\'echat-Haefliger invariant defined in
\cite[\S1.2, \S2.2]{CSI})
is 1--1 and $\varkappa=\varkappa_{\# }q_{\# }$ is surjective:
$$\xymatrix{E^7(N) \ar[r]^{q_{\# }} &  E_{\# }^7(N) \ar[r]^(0.15){\varkappa_{\# }} &
H_2^{DIFF}:=\{u\in H_2\ |\ \rho_2u=w_2^*(N),\ u\capM{N} u=\sigma(N)\}\subset H_2 }.$$
For each $u\in H_2^{DIFF}$ we have $|\varkappa^{-1}(u)|=\widehat u/\gcd(u,2)$.}%
\footnote{This is proven in \cite{Sk10, CS11} building on \cite{BH70}.}

Our second main result is a generalization of the above statement to non-simply-connected 4-manifolds.
The maps $\varkappa$, $\lambda$, $\beta_{u,l}$, and $\eta_{u,l,b}$, $\theta_{u, l, b}$ of Theorem \ref{t:gensm} below are defined in \cite[\S2.2, \S2.3]{CSI} and in \S\ref{s:defandplan-eta}.

\smallskip
{\bf Definition of $\capM{d}$.}
For a symmetric pair $(u,l)$ and $d:=\mathop{\fam0 div} u\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$ a bilinear map
$$
\capM{d}\colon \mathop{\fam0 coker}(2\rho_d\overline l)\times\mathop{\fam0 ker}(2\rho_d\overline l)\to{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d\quad\text{is well-defined by}
\quad [c]\capM{d} y:=c\capM{N} y.
\footnote{Indeed, for each $x\in H_3$ and $y\in \mathop{\fam0 ker}(2\rho_d\overline l)$ we have
$2\overline lx\capM{N} y=2l(x,y)\underset d\equiv2l(y,x)=2\overline ly\capM{N} x\underset d\equiv0$.
Hence $\mathop{\fam0 im}(2\rho_d\overline l)\capM{N} \mathop{\fam0 ker}(2\rho_d\overline l)=\{0\}\subset{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$.}
$$

\begin{Theorem}\label{t:gensm}
Let $N$ be a closed connected orientable 4-manifold with torsion free $H_1$.

(a) The product
$$\varkappa \times\lambda\colon E^7(N)\to H_2^{DIFF}\times B(H_3)$$
has non-empty image consisting of symmetric pairs.

(b) For each $(u,l)\in \mathop{\fam0 im} (\varkappa\times\lambda)$ denote $d:=\mathop{\fam0 div} u$.
Each map
$$
\beta_{u,l}\colon (\varkappa\times\lambda)^{-1}(u,l)\to\mathop{\fam0 coker}(2\rho_d\overline l)
$$
is surjective (see the remark immediately below the Theorem).

(c) For each $b\in \mathop{\fam0 coker}(2\rho_d\overline l)$ the map
$$
\theta_{u,l,b}\colon \mathop{\fam0 ker}(2\rho_d\overline l)\to4{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}
$$
is a homomorphism and each map
$$\eta_{u,l,b}\colon \beta_{u,l}^{-1}(b)\to\frac{{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}}{\mathop{\fam0 im}\theta_{u,l,b}}$$
is an injection whose image consists of all even elements (see the remark immediately below the Theorem).
Moreover,
$$
\theta_{u,l,b}(y)-\theta_{u,l,b'}(y)=4\rho_{\widehat d}(b-b')\capM{d} y
\quad\text{for each}\quad y\in \mathop{\fam0 ker}(2\rho_d\overline l)\subset H_3.
$$

(d) $|\beta_{u,l}^{-1}(b)|=
\dfrac{\widehat u}{\gcd(u,2)\cdot|\mathop{\fam0 im} \theta_{u,l,b}|}.$
\end{Theorem}

{\bf Remark on relative invariants.}
We call geometrically defined maps invariants (this is informal, formally an invariant is the same as a map).
The maps $\lambda$ and $\varkappa$ are
invariants.
The maps $\beta_{u,l}$ and $\eta_{u, l, b}$ are {\em relative invariants}.
For $\eta_{u, l, b}$ this means that for $[f_0], [f_1] \in \beta_{u,l}^{-1}(b)$
there is an
invariant $([f_0], [f_1]) \mapsto \eta(f_0, f_1)$ (defined in \S\ref{s:defandplan-eta})
and that $\eta_{u, l,b}(f) := \eta(f,f')$ for a fixed choice of $[f']\in\beta_{u,l}^{-1}(b)$.
We
suppress the choice of $[f']$ from the notation.
For $\beta_{u,l}$ the situation is similar.

\smallskip
Parts (a) and (b) of Theorem \ref{t:gensm}
follow from \cite[Theorem 1.2]{CSI}.
The new part of Theorem \ref{t:gensm} is (c), which is proven in in \S\ref{s:defandplan-eta}.
Part (d) follows  because by (c) $\mathop{\fam0 im}\eta_{u,l,b}=2{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}/\mathop{\fam0 im} \theta_{u,l,b}$.

We remark that Theorem \ref{t:s1s3sm} is not an immediate corollary of Theorem \ref{t:gensm},
cf.~\cite[Remarks 1.4.a and 2.16]{CSI}.

Identify $E^7(S^4)$ and ${\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$ by any isomorphism.

\begin{Addendum}
\label{l:addeta}
In the notation of Theorem \ref{t:gensm}\remdc{,} for each $a\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$ and $f\in\beta_{u,l}^{-1}(b)$
$$\eta_{u,l,b}(f\#a)=\eta_{u,l,b}(f)+[2a]\in \frac{{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}}{\mathop{\fam0 im} \theta_{u,l,b}}.$$
\end{Addendum}

This follows from the definition of $\eta_{u,l,b}$ (\S\ref{s:defandplan-eta}) and \cite[Lemma 4.3.b]{CSI}.

\begin{Corollary}\label{c:gen}
For each $(u,l)\in \mathop{\fam0 im} (\varkappa\times\lambda)$ let $d:=\mathop{\fam0 div} u$.
There is $f_{u,l}\in (\varkappa\times\lambda)^{-1}(u,l)$ such that for each $f\in (\varkappa\times\lambda)^{-1}(u,l)$ and $a,a'\in {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$, denoting $b:=\beta(f,f_{u,l})\in\mathop{\fam0 coker}(2\rho_d\overline l)$ we have

(a) \quad $f\#a=f\#a'\quad\Leftrightarrow\quad a=a'$, \quad provided

$\bullet$ $u=0$ and $\mathop{\fam0 div} b$ is divisible by 6, or

$\bullet$ $u\ne0$, $2\rho_d\overline l=0$ and $u$ is divisible by $24\mathop{\fam0 ord}(4b)$;

(b) \quad $f\#a=f\#a'\quad\Leftrightarrow\quad a\equiv a'\mod2\gcd(\mathop{\fam0 div} b,6),$ \quad provided $u=0$;

(c) \quad $f\#a=f\#a'\quad\Leftrightarrow\quad a\equiv a'\mod \dfrac{\widehat{\frac u{\mathop{\fam0 ord}(4b)}}}{\gcd(u,2)},$ \quad
provided $u\ne0$ and $2\rho_d\overline l=0$.
\end{Corollary}

The class $u$ is divisible by $d$ and hence by the order $\mathop{\fam0 ord}(4b)$ of $d$ in
the $d$-group $\mathop{\fam0 coker}(2\rho_d\overline l)$.

Part (a) follows from Parts (b,c).
Parts (b,c) are proven in \S\ref{s:defandplan-s1s3d}.
Cf.~Remark \ref{r:cordif} below.

\begin{Corollary} \label{r:s1s3sm}
Theorem \ref{t:gensm} has the following restatement analogous to Theorem \ref{t:s1s3sm} and to
\cite[Corollary 2.14.b]{CSI}.
There is a surjection
$$\tau:{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}\times H_1\times H_2^{DIFF}\times B_0(H_3)\to E^7(N)\quad\text{such that}$$
$$\tau(a,b,u,l)=\tau(a',b',u',l')\quad\Leftrightarrow\quad
u=u',\quad l=l',\quad b-b'\in\mathop{\fam0 ker}(2\rho_{\mathop{\fam0 div} u}\overline{l_u})\quad\text{and}\quad a-a'\in\mathop{\fam0 im}\eta_{u,l_u,b},$$
where $l_u:=l+\lambda\tau(0,0,u,0)$.
\end{Corollary}

\comment

\begin{Corollary}\label{c:gen} Assume the notation and the hypothesis of Theorem \ref{t:gensm}.
For each $(u,l)\in \mathop{\fam0 im} (\varkappa\times\lambda)$ we may choose $\beta_{u,l}$ so that for each $b\in \mathop{\fam0 coker}(2\rho_d\overline l)$

(a) $|\beta_{u,l}^{-1}(b)|=2$ when $u$ is divisible by 4 but neither by 8 nor by 3.

(b) $|\beta_{u,l}^{-1}(b)|=1$ if $u$ is divisible neither by 4 nor by 3.

(c) $|\beta_{u,l}^{-1}(b)|=12$ if either

$\bullet$ $u=0$ and $b$ is divisible by 6, or

$\bullet$ $u\ne0$, $2\rho_d\overline l=0$ and $u$ is divisible by $24\mathop{\fam0 ord}(4b)$.

$$(d)\qquad |\beta_{u,l}^{-1}(b)|=\begin{cases} 2\gcd(\mathop{\fam0 div} b,6) &u=0\\
\dfrac{\widehat{\frac d{\mathop{\fam0 ord}(4b)}}}{\gcd(d,2)} &u\ne0 \text{ and }2\rho_d\overline l=0\end{cases}.$$
(Here we give no statement for $\beta_{u,l}^{-1}(b)$ in other cases.)
\end{Corollary}

Part (c) follows from (d)

\endcomment

\subsection{Discussion of the action of knots}\label{s:action}

\begin{Remark}[The action of knots in Theorem \ref{t:gensm}] \label{r:at}
(a) Take any $[f]\in E^7(N)$.
Let
$$
O(f)=O([f]):=\{[f\#g]\ :\ [g]\in E^7(S^4)\}
$$
be  the orbit of $[f]$ under the action of $E^7(S^4)$.
We have $O(f)=\beta_{u,l}^{-1}(b)$ when $[f]\in \beta_{u,l}^{-1}(b)$ by \cite[Theorem 1.2]{CSI} and
the additivity of $\varkappa,\lambda$ and $\beta$ \cite[Lemmas 2.3 and 2.9]{CSI}.

Define the {\it inertia group} of $f$, $I(f) \subset E^7(S^4)={\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$,%
\footnote{The inertia group of $f$ is just the stabilizer of $[f]$ under the action of $E^7(S^4)$.
We use the word `inertia' following its use for the action of the group homotopy spheres on
the diffeomorphism classes of smooth manifolds: see the second paragraph of Remark \ref{r:general}.}
to be the subgroup of isotopy classes in $E^7(S^4)$ which do not change
$[f]$ after embedded connected sum:
$$
I(f)=I([f])  := \{ [g]\in E^7(S^4) \, : \, [f \#g]=[f]\}
$$
For some cases this
orbit and group are found in terms of $u,l,b$ in Corollaries \ref{t:actioni} and \ref{c:gen}.

(b) Problem: characterize those $f$ for which $|O(f)|=12$ (i.e.~$|I(f)|=1$),
and those $f$ for which $|O(f)|=1$ (i.e.~$|I(f)|=12$).

(c) The indeterminancy in the classification of Theorem \ref{t:gensm}.c corresponds to the fact that we do not always know  $\mathop{\fam0 im}\theta_{u,l,b}$.
Thus determining $\mathop{\fam0 im}\theta_{u, l, b}$ becomes a key problem.
This image is found in this paper when either $u=0$ or $2\rho_d\overline l=0$ (Corollary \ref{c:gen}) or in the cases (1,2,3) of Remark \ref{r:cordif} below.
For general $u,l$ (and even simple enough $N$) there are some $b$ for which the methods of this paper do not completely determine $\mathop{\fam0 im}\theta_{u,l,b}$.
\end{Remark}

\begin{Remark}[The action of knots in general] \label{r:general}
If the quotient $E_{\# }^m(P)$ is known for a closed $n$-manifold $P$, the description of $E^m(P)$ is reduced
to the {\it determination of the orbits} of the embedded connected sum action of $E^m(S^n)$ on $E^m(P)$.
For a general closed $n$-manifold $P$ describing the  action by a non-zero group of knots
$E^m(S^n)$ on $E^m(P)$ is a non-trivial task.
For the cases when the quotient $E_{\# }^7(N)$ coincides with the set of PL embeddings up to PL isotopy, the quotient
has been known since 1960s \cite{MAE, MAF, MAT}.
However, until recently no description of the action (or, equivalently, no classification of $E^m(P)$) was known for $E^m(S^n)\ne0$ and $P$ not a disjoint union of homology spheres.
For recent results see \cite{Sk08'} and \cite{Sk10, CS11} mentioned above.
On the other hand, the description of the action in \cite{CRS07, Sk11, CRS12, Sk15} is not hard,
the hard part of the cases considered there is rather the description of the quotient $E_{\# }^7(N)$.

There are non-isotopic embeddings $g_1,g_2\colon S^2\to S^4$ and an embedding $f\colon\R P^2\to S^4$ such that $[f\#g_1]\ne[f\#g_2]$ \cite{Vi73}.
I.e. the action of the monoid $E^4(S^2)$ on $E^4(\R P^2)$ is not free.

Various authors have studied the analogous connected sum action of the group of homotopy $n$-spheres on the set of smooth $n$-manifolds homeomorphic to given manifold; see for example \cite{Sc73, Wi74} and references there.
\end{Remark}

\begin{Remark}[An approach to the action of knots] \label{r:action}
Let us explain the ideas required to move from the classification modulo knots in \cite{CSI} to the
main results of this paper.
We briefly recall and continue the discussion in \cite[1.4]{CSI}.

Suppose that $f_0, f_1 \colon N \to S^7$ are embeddings.
Assume that $f_1$ is isotopic to $f_0 \# g$ for some embedding $g \colon S^4 \to S^7$.
By \cite[Isotopy Classification Modulo Knots Theorem 2.8]{CSI} this is equivalent to $\lambda(f_0)=\lambda(f_1)$, $\varkappa(f_0)=\varkappa (f_1)$ and $\beta(f_0,f_1)=0$.
The complements $C_0$ and $C_1$ may be glued together along a bundle isomorphism
$\varphi\colon\partial C_0\to\partial C_1$ to form a parallelizable closed $7$-manifold $M = C_0 \cup_{\varphi} (-C_1)$.
Recall that $d: = \mathop{\fam0 div}\varkappa(f_0)$ is the divisibility of $\varkappa(f_0) \in H_2$.
By the assumption on $f_0,f_1$ there is a joint Seifert class $Y\in H_5(M)$ such that $\rho_dY^2=0$, i.e.~{\it a $d$-class} \cite[Lemma 4.1]{CSI}.
There is a spin null-bordism $(W, z)$ of $(M_\varphi, Y)$, since $\Omega_7^{Spin}(\C P^\infty) = 0$.
Since $\rho_dY^2 = 0$, the class $\rho_dz^2 \in H_4(W, \partial;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d)$ lifts to $\overline{z^2} \in H_4(W;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d)$.
Recall that $p^*_W \in H_4(W, \partial)$ is the Poincar\'{e} dual of $p_W$, the spin Pontrjagin class of $W$.
We then verified that {\it the Kreck invariant},
$$
\eta(\varphi,Y) := \overline{z^2}\cap_W\rho_{\widehat d}(z^2-p^*_W) \in {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d},
$$
determines the surgery obstruction for $W$ to be spin diffeomorphic to the product $C_0\times I$ \cite{CSI}.
We proved that $\eta(\varphi,Y)$ is independent of the choices of $W,z,\overline{z^2}$ for
a fixed bundle isomorphism $\varphi$ and $d$-class $Y$ \cite[\S4.1]{CSI}.
We also proved that $\eta(\varphi,Y)$ is independent of the choice of $\varphi$: for the precise statement, see
\cite[Lemma 4.3.c]{CSI}.
So we need to know the various values of $\eta(\varphi,Y)$ arising from the different possible choices of $Y$.
These choices are described in \cite[Description of $d$-classes for $M_f$ Lemma 4.7]{CSI}.
The achievement of this paper is showing that the change of $\eta(\varphi,Y)$ under a change of $Y$ is precisely determined by $\theta_{u, l, b}$, and proving the properties of $\theta_{u, l, b}$ (Lemma \ref{l:diff}).
\end{Remark}

\comment

\remas{delete?

and, when $I(f) \neq E^7(S^4)$, to find an invariant which detects the action of knots.
When $H_1 = 0$ we did this in \cite{CS11}.
We consider the case where $H_1 \neq 0$ by continuing the discussion of the knotting problem in \cite[1.4]{CSI}.

When $H_1\ne0$ the intersection form on $H_4(W)$ has an orthogonal splitting
\begin{equation} \label{eq:splitting}
H_4(W) \cong H_4(M) \oplus X
\end{equation}
where the intersection form on $X$ is unimodular and $H_4(M)$ is the radical of the intersection form.

Analysis of the modified surgery obstruction for $(W, z)$ shows that we must choose $Y$ so that $Y^2 = 0$.

In this case the intersection form on $X$ and $p_W$ allow us to compute $\eta(W,z)$
as above and so determine the isotopy class of $g$ modulo $I(f_0)$.

To apply the method above to determine the action of knots

In \cite[Description of $d$-classes for $M_f$ Lemma 4.7]{CSI} we prove that $\mathop{\fam0 ker}{2 \rho_d \overline l}$ acts freely on the set of $d$-classes.

For $f \in \beta_{u, l}^{-1}(b)$ we show that
\[ \mathop{\fam0 im}(\theta_{u, l, b}) = \{\eta(\varphi,Y) \, | \,  \partial(W, z) = (M, Y), Y^2 = 0 \} \]}

\remas{The following results are great but not proved so have to be postponed until next version.}

\begin{Corollary} \label{c:iss}
\remas{prove!} There are embeddings $f_0\colon S^1 \times S^3 \to S^7$ and $f_1 \colon S^2 \times S^2 \to S^7$
such that

$\bullet$ for each embedding $g:S^4\to S^7$ we have $[f_0\#g]=[f_0]$ and $[f_1\#g]=[f_1]$ but

$\bullet$ there is an embedding $g:S^4\to S^7$ such that $[f_0\#f_1\#g]\ne[f_0\#f_1]$.
\end{Corollary}

\begin{proof}
By Theorem \ref{t:s1s3sm}.a there are

$\bullet$ an embedding $f_0$ with $\lambda(f_0) = 4 \in B(H_3) = {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$.

$\bullet$ an embedding $f_1$ with $\varkappa(f_1) = (24, 0) \in H_2(S^2 \times S^2) = {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}^2$.

We have $\varkappa(f_0\# f_1)=u:=(24, 0)$ and $\lambda(f_0\# f_1)=l:=8$ \remas{prove!}.
Then $\mathop{\fam0 div} u = 24$ and $\mathop{\fam0 ker}(2 \rho_{\mathop{\fam0 div} u} \overline l) = 3H_3$.
By Theorem \ref{t:gensm}c we can find $b \in \mathop{\fam0 coker}(2 \rho_d \bar l)$ with $\theta_{u, l, b} \neq 0$ \remas{prove!}.
Then $|\beta_{u, l}^{-1}(b)|>1$.
\end{proof}

\begin{Corollary} \label{ex:is}
\remas{prove!}
For $j = 2, 6$, there are embeddings $f_0^j \colon S^1 \times S^3 \to S^7$ and $f_1 \colon S^2 \times S^2 \to S^7$
such that $I(f_0^j) = 0 = I(f_1)$ but $I(f_0^j \# f_1) \cong {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{j}$.
\end{Corollary}

\begin{proof}
For $f_0^j$ take any embedding with $\lambda(f_0^j) = 2j \in B(H_3) = {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$
and for $f_1$ take any embedding with $\varkappa(f_1) = (24, 0) \in H_2(S^2 \times S^2) = {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}^2$.
We have $(u, l) : = (\varkappa \times \lambda)(f_0^j \# f_1) = \bigl( (24, 0), 4j \bigr)$.
In this case $d = 24$ and $\mathop{\fam0 ker}(2 \rho_{24} \bar l) = (j/2) \cdot H_3$ \remas{wrong for $j=2$}.
By Theorem \ref{t:gensm}c we can find $b \in \mathop{\fam0 coker}(2 \rho_d \bar l)$ with $\theta_{u, l, b} \neq 0$ \remas{prove!}.
By Theorem \ref{t:gensm}c for all $f \in \beta_{u, l}^{-1}(b)$, $I(f) \cong j \cdot {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$.
\end{proof}

\begin{Remark}\label{r:inman}
Recall that the inertia group $I(P)$ of a closed oriented $n$-manifold $P$ is defined to the subgroup
of homotopy spheres $\Sigma \in \Theta_n$ such that there is an orientation preserving diffeomorphism
$P \cong P \# \Sigma$.
In particular, our Corollary \ref{ex:is} can be compared with \remas{\cite[\S6]{Wi74}} where examples of
$7$-manifolds $P_0$ and $P_1$ are given with $I(P_0) = I(P_1) = 0$, but $I(P_0 \# P_1) \ne 0$.
\end{Remark}

\endcomment

\section{Definition of invariants and proofs modulo Lemma \ref{l:diff}} \label{s:defplan}

\subsection{Definition of the $\eta$- and $\theta$-invariants and proof of Theorem \ref{t:gensm}.c}
\label{s:defandplan-eta}

In this paper we use the notation and definitions of \cite[\S\S 2.1-2.3, 4.1]{CSI}.
In particular, we assume that groups $H_q:=H_q(N; {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H})$ are torsion free.
Denote $\theta(f,y):=\eta(\mathop{\fam0 id}\partial C_f,Y_{f,y})\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}.$

\begin{Lemma}[proved in \S\ref{s:eta-prb},\S\ref{s:eta-prac}]\label{l:diff}
(a) $\theta(f,y)$ is divisible by 4 for each $f,y$.

(b) The map $\theta(f,\cdot):\mathop{\fam0 ker}(2\rho_d\overline{\lambda(f)})\to{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}$ is a homomorphism, where $d:=\mathop{\fam0 div}\varkappa(f)$.

(c) For each $(u,l)\in \mathop{\fam0 im} (\varkappa\times\lambda)$, \ $d:=\mathop{\fam0 div} u$, representatives $f_0,f_1$ of two isotopy classes in $(\varkappa\times\lambda)^{-1}(u,l)$ and $y\in\mathop{\fam0 ker}(2\rho_d\overline l)$ we have
$\theta(f_0,y)-\theta(f_1,y)=4\rho_{\widehat d}(\beta(f_0,f_1)\capM{N} y)$.
\end{Lemma}

{\bf Definition of $\theta_{u,l,b}$.}
Take any $(u,l)\in \mathop{\fam0 im} (\varkappa\times\lambda)$ and $b\in K_{u,l}$. Let $d:=\mathop{\fam0 div} u$.
Define
$$\theta_{u,l,b}:\mathop{\fam0 ker}(2\rho_d\overline l)\to4{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}\quad\text{by}\quad\theta_{u,l,b}(y):=
\theta(f,y), \quad\text{where}\quad [f]\in\beta_{u,l}^{-1}(b).$$
The map $\theta_{u,l,b}$ is well-defined (i.e.~is independent of the choice of $f$) and is a homomorphism by Lemma \ref{l:diff}.ab and the transitivity of $\beta$ \cite[Lemma 2.10]{CSI}.

\smallskip
{\bf Definition of $\eta(f_0,f_1)$.}
Take representatives $f_0,f_1$ of two isotopy classes in $(\varkappa\times\lambda)^{-1}(u,l)$ such that $\beta(f_0,f_1)=0$.
By \cite[Lemma 2.5]{CSI}
there is a $\pi$-isomorphism $\varphi:\partial C_0\to\partial C_1$.
By \cite[Lemma 4.1]{CSI}
there is a $d$-class $Y\in H_5(M_\varphi)$.
Define
$$
\eta(f_0,f_1):=[\eta(\varphi,Y)]\in\frac{{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}}{\mathop{\fam0 im}\theta_{u,l,b}}.
$$
This is well-defined by \cite[Lemma 4.3.c]{CSI}
and Lemma \ref{l:welleta}.a below, and is even by \cite[Lemma 4.3.a]{CSI}.

\begin{Lemma}\label{l:traeta}
Let $f_0,f_1,f_2:N\to S^7$  be embeddings, $\varphi_{01}:\partial C_0\to\partial C_1$ and $\varphi_{12}:\partial C_1\to\partial C_2$
$\pi$-isomorphisms,
$Y_{01}\in H_5(M_{\varphi_{01}})$ and $Y_{12}\in H_5(M_{\varphi_{12}})$ \ $d$-classes.
Then $\varphi_{02}:=\varphi_{12}\varphi_{01}$ is a $\pi$-isomorphism and there is a $d$-class
$Y_{02}\in H_5(M_{\varphi_{02}})$ such that $\eta(\varphi_{02},Y_{02})=\eta(\varphi_{01},Y_{01})+\eta(\varphi_{02},Y_{12})$.
\end{Lemma}

This is proved analogously to \cite[Lemma 2.10]{CS11}, cf.~\cite[\S2, Additivity Lemma]{Sk08'} (the property that $Y_{02}$ is a $d$-class is achieved analogously to \cite[proof of Lemma 4.6]{CSI}).

\begin{Lemma}\label{l:welleta}
Take representatives $f_0,f_1$ of two isotopy classes in $(\varkappa\times\lambda)^{-1}(u,l)$ such that $\beta(f_0,f_1)=0$.
Denote $d:=\mathop{\fam0 div} u$.

(a) For each $\pi$-isomorphism $\varphi:\partial C_0\to\partial C_1$ the residue $\eta(f_0,f_1)$ is independent of the choice of a $d$-class $Y\in H_5(M_\varphi)$.

(b) If $\eta(f_0,f_1)=0$, then for each $\pi$-isomorphism $\varphi:\partial C_0\to\partial C_1$ there is a $d$-class
$Y\in H_5(M_\varphi)$ such that $\eta(\varphi,Y)=0\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}$.

\end{Lemma}

\begin{proof}[Proof of (a)]
Take any $d$-classes $Y',Y''\in H_5(M_\varphi)$.
Part (a) follows because
$$
\eta(\varphi,Y')-\eta(\varphi,Y'')\overset{(1)}= \eta(\mathop{\fam0 id}\partial C_0,Y)\overset{(2)}= \theta(f_0,y)=\theta_{u,l,\beta(f_0,f')}(y)\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d},
$$
where

$\bullet$ equality (1) holds for some $d$-class $Y\in H_5(M_{f_0})$ by Lemma \ref{l:traeta};

$\bullet$ equality (2) holds for some $y\in\mathop{\fam0 ker}(2\rho_d\overline l)$ by description of $d$-classes \cite[Lemma 4.7]{CSI}.
\end{proof}

\begin{proof}[Proof of (b)]
Take any $\pi$-isomorphism $\varphi:\partial C_0\to\partial C_1$.
Part (b) follows because
$$0\overset{(1)}=\eta(\varphi,Y')-\theta_{u,l,\beta_{u,l}(f_0)}(y)=\eta(\varphi,Y')-\theta(f_0,y)\overset{(3)}= \eta(\varphi,Y)\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d},$$
where

$\bullet$ equality (1) holds for some $d$-class $Y'\in H_5(M_\varphi)$ and $y\in\mathop{\fam0 ker}(2\rho_d\overline l)$ because $\eta(f_0,f_1)=0$;

$\bullet$ equality (3) holds for some $d$-class $Y\in H_5(M_\varphi)$ by Lemma \ref{l:traeta}.
\end{proof}

\begin{Lemma}[Transitivity of $\eta$]\label{l:difeta}
For each three embeddings $f_0,f_1,f_2:N\to S^7$ having the same values of $\varkappa$- and $\lambda$-invariants
and such that $\beta(f_0,f_1)=\beta(f_1,f_2)=0$ we have $\eta(f_2,f_0)=\eta(f_2,f_1)+\eta(f_1,f_0)$.
\end{Lemma}

This follows by Lemma \ref{l:traeta}.

\begin{Theorem}[Isotopy classification]\label{l:Is}
If $\lambda(f_0)=\lambda(f_1)$, $\varkappa(f_0)=\varkappa (f_1)$, $\beta(f_0,f_1)=0$ and
$\eta(f_0,f_1)=0$, then $f_0$ is isotopic to $f_1$.
\end{Theorem}

\begin{proof} Analogously to the proof of \cite[Isotopy Classification Modulo Knots Theorem 2.8]{CSI}.
Only replace the second paragraph by `Since $\eta(f_0,f_1)=0$, by Lemma \ref{l:welleta}.b we can change $Y$ and assume additionally that   $\eta(\varphi,Y)=0$.'
\end{proof}

{\bf Definition of $\eta_{u,l,b}$.}
Take any $[f_0]\in \beta_{u,l}^{-1}(b)$.
Define a map
$$
\eta_{u,l,b}:\beta_{u,l}^{-1}(b)\to\frac{{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}}{\mathop{\fam0 im}\theta_{u,l,b}}\quad\text{by}\quad \eta_{u,l,b}[f]:=\eta(f,f_0).
$$
The map $\eta_{u,l,b}$ depends on $f_0$ but we do not indicate this in the notation.

\begin{proof}[Proof of Theorem \ref{t:gensm}.c]
The property on $\theta_{u,l,b}-\theta_{u,l,b'}$ holds by Lemma \ref{l:diff}.c.
The map $\eta_{u,l,b}$ is injective by the Isotopy Classification Theorem \ref{l:Is}.
The image of this map consists of all even elements by \cite[Lemma 4.3.a]{CSI}
and Addendum \ref{l:addeta}.
\end{proof}

\comment

\begin{Lemma}[Transitivity of $\beta$]\label{l:Dif}
For every triple of embeddings $f_0,f_1,f_2:N\to S^7$ having the same values of $\varkappa$- and $\lambda$-invariants
we have $\beta(f_2,f_0)=\beta(f_2,f_1)+\beta(f_1,f_0)$.
\cite[Lemma 2.10]{CSI}
\end{Lemma}

\begin{Lemma}\label{l:piso}
If $\varkappa(f_0)=\varkappa(f_1)$ and $\lambda(f_0)=\lambda(f_1)$, then there is a $\pi$-isomorphism
$\varphi:\partial C_0\to\partial C_1$.
A $\pi$-isomorphism is unique (up to vertical homotopy through linear isomorphisms) over $N_0$.
\cite[Lemma 2.5]{CSI}
\end{Lemma}

\begin{Lemma} \label{l:fc}
If $\lambda(f_0)=\lambda(f_1)$,\ $\varkappa(f_0)=\varkappa(f_1)$,\ $\beta(f_0,f_1)=0$ and
$\varphi\colon \partial C_0\to\partial C_1$ is a $\pi$-isomorphism, then there is a $\mathop{\fam0 div}(\varkappa(f_0))$-class for $\varphi$.
\cite[Lemma 4.1]{CSI}
\end{Lemma}

\begin{Lemma}[Description of $d$-classes for $M_f$] \label{l:simpleY0}
A class $Y\in H_5(M_f)$ is a $d$-class if and only if $Y=Y_{f,y}$ for some
$y\in \mathop{\fam0 ker}(2\rho_{\mathop{\fam0 div}(\varkappa(f))}\overline{\lambda(f)})$. \cite[Lemma 4.7]{CSI}
\end{Lemma}

\begin{Lemma}\label{l:etaeven} \cite[Lemma 4.3]{CSI}
Let $\varphi\colon \partial C_0\to \partial C_1$ be a $\pi$-isomorphism and $Y\in H_5(M_\varphi)$ a $d_0$-class.

(a) (Divisibility of $\eta$ by 2) The residue $\eta(\varphi,Y)\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat{d_0}}$ is even.

(b) (Change of $\eta$)
There is an embedding $g\colon S^4\to S^7$, a $\pi$-isomorphism $\varphi'\colon \partial C_0\to \partial C_{f_1\#g}$ and a $d_0$-class
$Y'\in H_5(M_{\varphi'})$ such that $\eta(\varphi',Y',f_0,f_1\#g,d_0)=\eta(\varphi,Y,f_0,f_1,d_0)+2$.

(c)(Change of $\varphi$)
For every $\pi$-isomorphism $\varphi'\colon \partial C_0\to \partial C_1$ there is a $d_0$-class $Y'\in H_5(M_{\varphi'})$ such that $\eta(\varphi',Y')=\eta(\varphi,Y)$.
\end{Lemma}

\endcomment

\subsection{Proof of
Corollary \ref{c:gen}.bc}
\label{s:defandplan-s1s3d}

\begin{Remark} \label{r:cordif}
In Corollary \ref{c:gen}.bc the assumption `$d=0$ or $2\rho_d\overline l=0$' can be replaced by each of the following successively weaker assumptions

(1) $\rho_{\widehat d}\mathop{\fam0 ker}(2\rho_d\overline l)\subset\rho_{\widehat d}H_3$ is a direct summand, or

(2) each homomorphism $\rho_{\widehat d}\mathop{\fam0 ker}(2\rho_d\overline l)\to4{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}$ extends to
$\rho_{\widehat d}H_3$, or

(3) there is an element $\widetilde b\in\mathop{\fam0 coker}(2\rho_d\overline l)$ such that $\theta_{u,l,\widetilde b}=0$.
\end{Remark}

Clearly,
$`\text{either }d=0\text{ or }2\rho_d\overline l=0'\ \Rightarrow\ (1)\ \Rightarrow\ (2).$

Denote $X_{u,l}:=\mathop{\fam0 ker}(2\rho_{\mathop{\fam0 div}(u)}\overline l)\subset H_3$.

\begin{proof}[Proof of $(2)\Rightarrow(3)$]
Take any $b'\in\mathop{\fam0 coker}(2\rho_d\overline l)$.
We have $\theta_{u,l,b'}=\theta_{u,l,b'}^+\rho_{\widehat d}$ for some homomorphism
$\theta_{u,l,b'}^+:\rho_{\widehat d}X_{u,l}\to 4{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}$.
Extend $\theta_{u,l,b'}^+$ to a homomorphism $\rho_{\widehat d}H_3\to 4{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}$.
Since $H_3$ is free, $\rho_{\widehat d}H_3$ is a free ${\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}$-module.
Hence the latter homomorphism is divisible by 4.
Then by Poincar\'e duality there is a class $x\in \rho_{\widehat d}H_1$ such that $\theta_{u,l,b'}^+(z)=4x\capM{N} z$
for each $z\in \rho_{\widehat d}X_{u,l}$.
Let $\widetilde b:=b'+[\widetilde x]$, where $\widetilde x\in \rho_dH_1$ is a lifting of $x$.
Then by Theorem \ref{t:gensm}
$$\theta_{u,l,\widetilde b}(y)=\theta_{u,l,b'}(y)-4\rho_{\widehat d}([\widetilde x]\capM{N} y)=
\theta_{u,l,b'}^+(\rho_{\widehat d}y)-4x\capM{N}\rho_{\widehat d}y =0
\quad\text{for each}\quad y\in X_{u,l}.$$
\end{proof}

\begin{proof}[Proof of the formula of Corollary \ref{c:gen}.bc under the assumption (3) of Remark \ref{r:cordif}]
Define $\beta_{u,l}'(f):=\widetilde b-\beta_{u,l}(f)$.
Then $\theta'_{u,l,b}=\theta_{u,l,\widetilde b-b}$ for each $b\in\mathop{\fam0 coker}(2\rho_d\overline l)$, hence $\theta'_{u,l,0}=0$.
Therefore we may assume that $\beta_{u,l}$ is chosen so that $\theta_{u,l,0}=0$.

Take any $b\in\mathop{\fam0 coker}(2\rho_d\overline l)$ and denote $X_{b,u,l}:=4b\capM{d} X_{u,l}$.
So
$$\gcd(d,2)\cdot|\beta_{u,l}^{-1}(b)|\overset{(1)}=
\frac{\widehat d}{|\mathop{\fam0 im}\theta_{u,l,b}|}\overset{(2)}=
\frac{\widehat d}{|\rho_{\widehat d}X_{b,u,l}|}=
[{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{\widehat d}:\rho_{\widehat d}X_{b,u,l}]\overset{(4)}=\gcd(\widehat d,[{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d:X_{b,u,l}])=
\widehat{[{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d:X_{b,u,l}]},$$
where equalities (1) and (2) hold by Theorem \ref{t:gensm}.
Now the formula of Corollary \ref{c:gen}.bc is implied by the following Lemma \ref{l:unimzd}.
\end{proof}

\begin{Lemma}\label{l:unimzd}
Let $V$ be a free ${\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$-module, $d$ an integer and $m:V\to V^*$ a homomorphism whose polarization $V\times V\to{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$
has a symmetric mod $d$ reduction.
Then for each $c\in\mathop{\fam0 coker}(\rho_d m)$
$$[{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d:c(\mathop{\fam0 ker}(\rho_dm))]=
\begin{cases} \mathop{\fam0 div} c &d=0\\
\dfrac d{\mathop{\fam0 ord} c} &d\ne0\end{cases}$$
($c(\mathop{\fam0 ker}(\rho_dm))\subset{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$ is defined analogously to definition of $\capM{d}$ before Theorem \ref{t:gensm}).
\end{Lemma}

\begin{proof}[Proof for $d=0$]
We need to prove the following.

{\it Let $V$ be a free ${\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$-module and $m:V\to V^*$ a homomorphism whose polarization $V\times V\to{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$ is symmetric.
Then for each $c\in V^*$ we have $[{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}:c(\mathop{\fam0 ker} m)]=\mathop{\fam0 div} c$.}

Let us prove the `$\subset$' part.
Assume that $q$ is a divisor of $c+\mathop{\fam0 Tors}\mathop{\fam0 coker} m$.
Then
there exist $s,l_1,\ldots,l_s\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$ and $c_0,t_1,\ldots,t_s\in V^*$ such that $l_nt_n\in \mathop{\fam0 im} m$ for each
$n=1,\ldots,s$ and $c=qc_0+t_1+\ldots+t_s$.
Take any $y\in\mathop{\fam0 ker} m$.
Since the polarization of $m$ is symmetric, we have $l_nt_n(y)=0\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$, thus $t_n(y)=0$.
Hence $c(y)=qc_0(y)+(t_1+\ldots+t_s)(y)=qc_0(y)$  is divisible by $q$.

Let us prove the `$\supset$' part.
Assume that $q$ is a divisor of $c|_{\mathop{\fam0 ker} m}$.
The subgroup $\mathop{\fam0 ker} m\subset V$ is a direct summand.
Take a decomposition $V=\mathop{\fam0 ker} m\oplus V'$.
Since $m|_{V'}:V'\to\mathop{\fam0 im} m$ is an isomorphism, there is an element $x\in V'$ such that $c|_{V'}=m(x)|_{V'}$.
Since the polarization of $m$ is symmetric, $m(x)|_{\mathop{\fam0 ker} m}=0$.
Then  $c-m(x)$ coincides with $c$ on $\mathop{\fam0 ker} m$ and is zero on $V'$.
So $c-m(x)=qc_0$ for some $c_0\in V^*$.
Hence $d(c+\mathop{\fam0 Tors}\mathop{\fam0 coker} m)$ is divisible by $q$.
\end{proof}

\begin{proof}[Proof for $d\ne0$]
We need to prove that {\it $|c(\mathop{\fam0 ker}(\rho_dm))|=\mathop{\fam0 ord} c$ for each
$c\in \mathop{\fam0 coker}(\rho_dm)$.}
(We remark that this is obvious for $\rho_dm=0$, which case is sufficient for Corollary \ref{c:gen}.c.)

Denote $K:=\rho_d\mathop{\fam0 ker}(\rho_dm)\subset V/dV$.
Since the polarization $V\times V\to{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$ of $m$ has a symmetric mod $d$ reduction,
$\mathop{\fam0 im}(\rho_d m)\subset K^\perp\subset V^*/dV^*$.
Since $|\mathop{\fam0 im}(\rho_d m)|=\dfrac{|V/dV|}{|K|}=|K^\perp|$, it follows that $\mathop{\fam0 im}(\rho_d m)=K^\perp$.
Now the required assertion follows because for each $c'\in V^*/dV^*$
$$|c'(K)|=\frac d{\mathop{\fam0 div} c'(K)}= \min\{r\ |\ rc'(K)=0\}=
\min\{r\ |\ rc'\in K^\perp\}=\mathop{\fam0 ord}\phantom{}_{(V^*/dV^*)/K^\perp}(c'+K^\perp).$$
\end{proof}

\begin{Addendum} \label{c:s1s3}
For each $l\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}-\{0\}$ there is a map
$\psi_l:{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}\times{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{2l}\to{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$ such that for each $a,a'\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$ and $l,b,l',b'\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$ we have  $a\#\tau(l,b)=a'\#\tau(l',b')$ if and only if
$$
\left[\begin{matrix}\text{either}\quad l=l'=0, \quad b=b'\quad\text{and}\quad a\equiv a'\mathop{\fam0 mod} 2\gcd(b,6)\\
\text{or}\quad l=l'\ne0, \quad b\equiv b'\mo2l \quad\text{and}\quad
\rho_{12}(a-a')=\psi_l([b/2l],\rho_{2l}b)-\psi_l([b'/2l],\rho_{2l}b) \end{matrix}\right.
$$
\end{Addendum}

\begin{proof}
By Theorem \ref{t:s1s3sm}.b if either $l\ne l'$ or $b\not\equiv b'\mo2l$, then the
equivalence is
clear because none of the two assertions holds.

Assume that $l=l'$ and $b\equiv b'\mo2l$.
Let $\tau:=a\#\tau(l,b)$ and $\tau'=a'\#\tau(l,b')$.
By the Isotopy Classification Theorem \ref{l:Is} and Theorem \ref{t:s1s3sm}.b
$\tau=\tau'\quad\Leftrightarrow\quad\eta(\tau,\tau')=0$.
Since $\mathop{\fam0 div}(\varkappa(\tau))=0$, we may use Corollary \ref{c:gen}.b.

If $l=0$, then $b=b'$.
By Theorem \ref{t:s1s3sm}.b and Corollary \ref{c:gen}.b $\mathop{\fam0 im}\theta_{0,0,b}$ is formed by elements of ${\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{24}$
divisible by $4\gcd(b,6)$.
Hence by Addendum \ref{l:addeta}
and the transitivity of $\eta$ (Lemma \ref{l:difeta})
$\eta(\tau,\tau')=\rho_{4\gcd(b,6)}(2a-2a')\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{4\gcd(b,6)}$.

If $l\ne0$, then by Theorem \ref{t:s1s3sm}.b and Corollary \ref{c:gen}.b $\mathop{\fam0 im}\theta_{0,l,b}=0$ and
$\eta(\tau,\tau')\in 2{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{24}$.
For each $x\in\{0,1,\ldots,2l-1\}$ and $k\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$ define $\psi_l(k,\rho_{2l}x):=\frac12\eta(\tau(l,x),\tau(l,x+2kl))\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$.
Then by Addendum \ref{l:addeta} and the transitivity of $\eta$ (Lemma \ref{l:difeta})
$$\eta(\tau,\tau')=
\rho_{24}(2a-2a') - \eta(\tau(l,\overline b),\tau(l,b)) + \eta(\tau(l,\overline b),\tau(l,b'))=
\rho_{24}(2a-2a') - \psi_l([\frac b{2l}],\rho_{2l}b) + \psi_l([\frac{b'}{2l}],\rho_{2l}b),$$
where $\overline b\in\{0,1,2,\ldots,2l-1\}$ is uniquely defined by $b\equiv \overline b\mo2l$.

These two formulas for $\eta(\tau,\tau')$ imply the equivalence describing preimages of $\tau$.
\end{proof}

\comment

\begin{Lemma}[Calculation]\label{l:calc}
For each integers $l,b,b'$ we have
$$\lambda(\tau(l, b)) = l\quad\text{and}\quad
\beta(\tau(l,b'),\tau(l,b))=\rho_{2l}(b-b')[S^1\times1_3]\in H_1(S^1\times S^3;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{2l})$$
\cite[the Calculation Lemmas 1.3.b, 2.7 and remark at the end of \S2.2]{CSI}.
\end{Lemma}

\begin{proof}[Proof of Theorem \ref{t:s1s3sm}: the surjectivity of $\tau$]
Take any $f\in E^7(S^1\times S^3)$.
Analogously to \cite[\S2.5, Proof of Theorem 1.1: the surjectivity of $\tau$]{CSI} we define $l,b$ and prove that $\beta(f,\tau')=0$ for $\tau':=\tau(0,l,b)$.
Therefore we can take $f'=\tau'$ in the definition of map $\beta_{u,l}$ and define map $\eta:=\theta_{0,l,b}$ and
residue $\eta(f,\tau')\in {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{24}/\mathop{\fam0 im}\eta$.
By the divisibility of $\eta$ by 2 \cite[Lemma 4.3.a]{CSI}
there is $a\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}$ such that $2a\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{24}$ is a lift of $-\eta(f,\tau')$.
By the additivity  and the transitivity of $\beta$ (Lemmas \ref{l:Addl}, \ref{l:Add} and \cite[Lemma 2.10]{CSI})
$$\lambda(f)=l=\lambda(\tau')=\lambda(\tau(a,l,b)),\quad \beta(f,\tau(a,l,b))=\beta(f,\tau')=0\quad$$
$$\text{and}\quad \eta(f,\tau(a,l,b))=\eta(f,\tau')+\eta(\tau',\tau(a,l,b))=\eta(f,\tau')-(a+\mathop{\fam0 im}\eta)=0.$$
Hence by the Isotopy Classification Theorem \ref{l:Is} $f=\tau(a,l,b)$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{t:s1s3sm}: the properties of $\tau({\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{12}\times(l,b))$]
It remains to prove the third property, cf.~the text after Theorem \ref{t:s1s3sm}.
\end{proof}

\endcomment

\section{Proof of Lemma \ref{l:diff} on $\theta$-invariant}\label{s:eta-pr}

In this section we use homological Alexander duality and the restriction homomorphism defined in \cite[\S3.1]{CSI}.

\subsection{Idea of proof of Lemma \ref{l:diff}}

We start with a lemma allowing to simplify the proof of the main result for $N=S^1\times S^3$ (Theorem \ref{t:s1s3sm}).
The simplified proof is not presented, so the lemma is not used in the sequel.
The standard embedding $\tau_0:S^1\times S^3\to S^7$ is defined in \cite[\S2.4]{CSI}.

\begin{Lemma}\label{l:etaiy} $\eta(\tau_0,y)=0$ for each $y\in H_3(S^1\times S^3)$.
\end{Lemma}

\begin{proof}
Define an extension
$$\mathop{\fam0 i}:D^2\times D^4\to S^7\quad\text{of $\tau_0$ by}\quad \mathop{\fam0 i}(x,y):=(y\sqrt{2-|x|^2},0,0,x)/\sqrt2.$$
Take an embedding $v_0:S^5\to S^7-\mathop{\fam0 i}(S^1\times D^4)$ whose linking coefficient with $\mathop{\fam0 i}(S^1\times D^0)$ is $y\capM{S^1\times S^3}[S^1\times1_3]$.
We omit subscript $\tau_0$ in this proof.
Since $\mathop{\fam0 lk}(\widehat Ay,\tau_0(S^1\times1_3))=y\capM{S^1\times S^3}[S^1\times1_3]$, we have
$\widehat Ay=[v_0(S^5)]\in H_5(C)\cong{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$.
We also have $\widehat Ay=i_{C}\nu^!y$.
Take a representative $P$ of $y$ and a chain
$$
V\in C_6(C)\quad\text{such that}\quad \partial V=\nu^{-1}P-v_0(S^5).
$$
Since $C$ is parallelizable, $v_0$ extends to an orientation-preserving embedding
$v_2:S^5\times D^2\to\mathop{\fam0 Int} C=\mathop{\fam0 Int} C\times\frac12$ transversal to $V$ and such that $\mathop{\fam0 im} v_2\cap V=v_0(S^5)$.
Extend $v_2$ to an orientation-preserving embedding $v_3:S^5\times D^3\to \mathop{\fam0 Int}(C\times I)$.
Let
$$
W_-:= C\times I-\mathop{\fam0 Int}\mathop{\fam0 im} v_3\quad\text{and}\quad W:=W_-\cup_{v_3|_{S^5\times S^2}}D^6\times S^2.$$
Consider the cohomology exact sequence of pair $(W,W_-)$ in the following Poincar\'e dual form
(analogous to the sequence (*) in \cite[Proof of Lemma 4.8]{CSI}):
$$\xymatrix{
H_6(D^6\times S^2
) \ar[r]  & H_6(W,\partial) \ar[r]^{r_{W_-}} & H_6(W_-,\partial) \ar[r] & H_5(
D^6\times S^2) \\
H^2(W,W_-)\ar[u]_{\cong}^{PD\circ\mathop{\fam0 ex}}  & & & H^3(W,W_-)\ar[u]_{\cong}^{PD\circ\mathop{\fam0 ex}}
}.$$
Since $H_5(D^6\times S^2)=0$, the map $r_{W_-}$ is an epimorphism.
Take any
$$Z\in r_{W_-}^{-1}(A[N]\times I\cap W_-)\subset H_6(W,\partial).$$
Denote
$$\widehat V:=V\cup D^6\times1_2\quad\text{and}\quad z:=Z+[\widehat V]\in H_6(W,\partial).$$
Since $H_5(D^6\times S^2)=0$, the spin structure on $W_-$ coming from $S^7\times I$ extends to $W$.
Clearly, $\partial W\underset{spin}=\partial(C\times I)\underset{spin}=M$
(for the `boundary' spin structure on $M$ coming from $C\times I$).
Since
$$\partial_WZ=\partial_{C\times I}(A[N]\times I)=Y_0\quad\text{and}\quad
\partial_W[\widehat V]=[\nu^{-1}P\times\frac12]=i_M\widehat Ay,
\quad\text{we have}\quad \partial_W z=Y_y.$$
By \cite[Lemma 4.7]{CSI}
$\partial_Wz^2=Y_y^2=0$.
So $z^2\in\mathop{\fam0 im} j_W$.
Analogously to (*) we obtain an isomorphism $H_4(C\times I)\cong H_4(W)$ commuting with
$i_{C\times I}:H_4(M)\to H_4(C\times I)$ and $i_W:H_4(M)\to H_4(W)$.
Since $i_{C\times I}$ is onto, $i_W$ is onto.
Hence $j_W=0$.
Thus $z^2=0$.
So take $\overline{z^2}:=0$ and obtain $\eta(\tau_0,y)=\overline{z^2}\capM{W}(z^2-p_W^*)=0$.
(We essentially proved that if $\widehat A_fy$ is spherical, then $\eta(f,y)=0$.)
\end{proof}

\subsection{Proof of Lemma \ref{l:diff}.b}\label{s:eta-prb}

In this and the following subsection $f,f_0,f_1:N\to S^7$ are embeddings representing any elements of $(\varkappa\times\lambda)^{-1}(u,l)$; we denote $d:=\mathop{\fam0 div}(u)$.

\smallskip
{\bf Definition of $W',W'_-$ and $i':W'\to W$.}
Let
$$W'_-:= C_f-\mathop{\fam0 Int}\mathop{\fam0 im} v_2,\quad W':=W'_-\cup_{v_2|_{S^2\times S^3\times S^1}}S^2\times D^4\times S^1,$$
(Manifold $W'$ may be called the result of an $S^1$-parametric surgery along $v_2$.)
Define an embedding $W'_-\to W_-$ by $x\mapsto x\times1/2$.
We assume that this embedding and the standard embedding $S^2\times D^4\times S^1\to S^2\times D^4\times S^2$
(that is the product of the identity and the equatorial inclusion $S^1\to S^2$) fit together to give an embedding $$i':W'\to W.$$
Observe that $\Delta,\widehat V\subset W'$.

\begin{Lemma}\label{l:resa}
For each $y\in H_3$ and $W_-,z,Z,V$ defined in \cite[Proof of Lemma 4.8]{CSI}
we have
$$
z^2\capM{W} W_-\underset d\equiv2i_{V,W_-}(Z\cap V)\in H_4(W_-,\partial)
$$
(since $\partial V\subset\partial W_-$, the inclusion induces a map $i_{V,W_-}:H_4(V,\partial)\to H_4(W_-,\partial)$).
\end{Lemma}

\begin{proof}
Since $\widehat V\subset W'$, we have $[\widehat V]^2=0\in H_4(W,\partial)$.
Also
$$Z^2\cap W_-=(A_f[N]\times I)^2\cap W_-=A_f\varkappa(f)\times I\cap W_-\underset d\equiv0\in H_4(W_-,\partial).$$
Hence
$$z^2\cap W_-=(Z+[\widehat V])^2\cap W_- \underset d\equiv 2(Z\capM{W}[\widehat V])\cap W_-=
2i_{V,W_-}(Z\cap\widehat V\cap W_-)=2i_{V,W_-}(Z\cap V).$$
\end{proof}

\begin{proof}[Proof of Lemma \ref{l:diff}.b]
In this proof a statement or a construction involving $k$ holds or is made for each $k=0,1$.
Given $y_k\in \mathop{\fam0 ker}(2\rho_d\overline l)$ construct manifold $W_k$ as $W$ of \cite[Proof of Lemma 4.8]{CSI}
by parametric surgery in $C_f\times[k-1,k]$.
We add subscript $k$ to $W_-,W'_-,t,\Delta,Z,\widehat V,z$ constructed in \cite[Proof of Lemma 4.8]{CSI}.
(So unlike in other parts of this paper, subscript 0 of a manifold does not mean deletion
of a codimension 0 ball from the manifold.)
Define
$$
W := W_0\cup_{C_f\times0}W_1\quad\text{and}\quad
W_-:=C_f\times[-1,1]-\mathop{\fam0 Int}\mathop{\fam0 im}(v_{3,0}\sqcup v_{3,1})=W_{0-}\cup_{C_f\times0}W_{1-}.
$$
This $W$  should not be confused with what were previously denoted $W$ but now is denoted $W_0$ and $W_1$.
Same remark should be done for $W_-$ and for $Z,V,\widehat V,z$ constructed below.

The spin structure on $W_-$ coming from $S^7\times[-1,1]$ extends to $W$.
Clearly, $\partial W\underset{spin}=$
\linebreak
$=\partial(C_f\times[-1,1])\underset{spin}\cong M_f$
(for the `boundary' spin structure on $\partial(C_f\times[-1,1])$ and on $M_f$).

Since $H_5(t_k\times\Delta_k)=0$, by the cohomological exact sequence of the pair $(W,W_-)$
(cf.~diagram (*) in \cite[Proof of Lemma 4.8]{CSI}), $r_{W_-}:H_6(W,\partial)\to H_6(W_-,\partial)$ is an epimorphism.
Take any
$$
Z\in r_{W_-}^{-1}(A_f[N]\times[-1,1]\cap W_-)\subset H_6(W,\partial).
$$
Denote
$$
V:=V_0\sqcup V_1,\quad \widehat V:=\widehat V_0\sqcup \widehat V_1\quad\text{and}\quad z:=Z+[\widehat V]\in H_6(W,\partial).
$$
Since $\partial_WZ=Y_{f,0}$ and $\partial_W[\widehat V_k]=i_{\partial W}\widehat A_fy_k$, we have $\partial z=Y_{f,y_0+y_1}$.
Thus the pair $(W,z)$ is a spin null-bordism of $(M_f,Y_{f,y_0+y_1})$.

Since $y_k\in\mathop{\fam0 ker}(2\rho_d\overline l)$, we have $\partial z_k^2\underset d\equiv0$.
Take any $\overline{z_k^2}\in j_{W_k}^{-1}\rho_dz_k^2$.
Let
$$
\overline{z^2}:=i_{W_0,W}\overline{z_0^2}+i_{W_1,W}\overline{z_1^2}.
$$
Then $\overline{z^2}\cap W_k=\overline{z_k^2}$.
Also
$$
j_W\overline{z^2}\cap W_- = \sum\limits_{k=0}^1j_{W_k}\overline{z_k^2}\cap W_- =
\rho_d\sum\limits_{k=0}^1z_k^2\cap W_{k-}\quad\text{and}
$$
$$
\sum\limits_{k=0}^1z_k^2\cap W_{k-}\overset{(1)}\equiv 2\sum\limits_{k=0}^1i_{V_k,W_{k-}}(Z_k\cap V_k) =
2i_{V,W_-}(Z\cap V) \overset{(3)}\equiv z^2\cap W_-.
$$
Here congruences (1) and (3) modulo $d$ hold by Lemma \ref{l:resa} and analogously to Lemma \ref{l:resa}, respectively.

Hence by the cohomological exact sequence of the pair $(W,W_-)$ with coefficients ${\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$
(cf.~diagram (*) in \cite[Proof of Lemma 4.8]{CSI})
$j_W\overline{z^2}-\rho_dz^2=n_0[t_0]+n_1[t_1]$ for some $n_0,n_1\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$.
We have
$$
n_k[t_k]=(j_W\overline{z^2}-\rho_dz^2)\cap W_k=j_{W_k}\overline{z_k^2}-\rho_dz_k^2 =0\in H_4(W_k,\partial;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d).
$$
Therefore $n_0=n_1=0$.
So $j_W\overline{z^2}=\rho_dz^2$.

Since $\widetilde{W_k}:=W_k-C_f\times[0,(2k-1)/3)$ is a deformation retract of $W_k$,
the inclusion $\widetilde{W_k}\to W_k$ induces an isomorphism on $H_4$.
Clearly, $z\cap W_k=z_k$, so $z^2\cap W_k=z_k^2$.
Hence
$$
\overline{z^2}\capM{W}(z^2-p_W)=\sum\limits_{k=0}^1 (\overline{z^2}\cap W_k)\capM{W_k}((z^2-p_W)\cap W_k)=
\sum\limits_{k=0}^1 \overline{z_k^2}\capM{W_k}(z_k^2-p_{W_k}).
$$
So $\eta(f,\cdot)$ is a homomorphism.
\end{proof}

\subsection{Proof of Lemma \ref{l:diff}.ac}\label{s:eta-prac}

\begin{Lemma}\label{l:res}
For each $y\in H_3$ and $W,Z,V,t,\Delta$ defined in \cite[Proof of Lemma 4.8]{CSI}
we have

(a) $\partial(Z\cap V)=[\partial\Delta]-i_{\partial C_f,\partial V}\xi y\in H_3(\partial V)$, where $\xi:N_0\to \partial C_f$
is  a weakly unlinked section for $f$ (see definition in \cite[\S2.2]{CSI}).

(b) $p_W=2m[t]\in H_4(W,\partial)$ for some $m\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$.
\end{Lemma}

Lemma \ref{l:res}.b is essentially proved in the proof of \cite[Lemma 4.8]{CSI}.

\begin{proof}[Proof of (a)]
The equality follows because
$$Z\cap V= (Z\cap W_-)\cap V= (A_f[N]\times I)\cap V= A_f[N]\cap V\in H_4(V,\partial)\quad\text{and}$$
$$\partial(A_f[N]\cap V)=A_f[N]\cap \partial V=
i_{\partial V}(A_f[N]\cap \mathop{\fam0 im} v)-i_{\partial V}(A_f[N]\cap\nu_f^{-1}P)\overset{(3)}= [\partial\Delta]-[\xi P].$$
Here $P$ and $v$ are defined in \cite[Proof of Lemma 4.8]{CSI}.
Equality (3) follows because

$\bullet$ $A_f[N]\cap\nu_f^{-1}P=[\xi P]$ by \cite[Lemma 3.2.a]{CSI}.

$\bullet$ $A_f[N]\cap \mathop{\fam0 im} v=[v(1_2\times S^3)]=[\partial\Delta]$ since
$$(A_f[N]\cap \mathop{\fam0 im} v)\capM{\mathop{\fam0 im} v}[v(S^2\times1_3)] = A_f[N]\capM{C_f} v(S^2\times1_3)\overset{(2)} = A_f[N]\capM{C_f}S^2_f=1.$$
Here equality (2) holds because $v(S^2\times1_3)$ is homologous to $S^2_f$ in $C_f$.
\end{proof}

\begin{Lemma}\label{l:prdiff2}
For each $y\in \mathop{\fam0 ker}(2\rho_d\overline l)$ and $W,W_-,z,t$ defined in \cite[Proof of Lemma 4.8]{CSI}
there is a class $\widehat{z^2}\in H_4(W;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d)$ such that

(a) $\overline{z^2}:=\widehat{z^2}+n[t]\in j_W^{-1}\rho_dz^2\subset H_4(W,\partial;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d)$ for some $n\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$.

(b) $[t]^2=(\widehat{z^2})^2=0\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$ and $[t]\capM{W}\widehat{z^2}=2\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$.
\end{Lemma}

The proof is given later in this section.
\footnote{Equality (3) from \cite[Proof of Lemma 4.3.a]{CSI}
also holds by \cite[Proof of Lemma 4.7]{CSI} and \ref{l:prdiff2} because
by Lemma \ref{l:prdiff2}.a for $d=2$ we have
$\eta'(\mathop{\fam0 id}\partial C_{f_0},Y_{f_0,y})= \overline{z^2}\capM{W} z^2=\overline{z^2}\capM{W}\overline{z^2} =2[t]\capM{W}\widehat{z^2}=0\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_2.$}

\begin{proof}[Proof of Lemma \ref{l:diff}.a]
The Lemma follows by \cite[Lemma 4.8]{CSI} and Lemmas
\ref{l:res}.b, \ref{l:prdiff2}.
Indeed,
$$\overline{z^2}\capM{W}(z^2-p_W^*)=\overline{z^2}\capM{W}\overline{z^2}-\overline{z^2}\capM{W} p_W^*\overset{(2)}= (\widehat{z^2}+n[t])^2-(\widehat{z^2}+n[t])\capM{W}2m[t]\overset{(3)}=4n-4m.$$
Here

$\bullet$ equality (2) holds by Lemma \ref{l:res}.b and property (a) of Lemma \ref{l:prdiff2},

$\bullet$ equality (3) holds by property (b) of Lemma \ref{l:prdiff2}.
\end{proof}

In the proof of Lemma \ref{l:diff}.c we will use not only the statement of Lemma \ref{l:prdiff2} but also
the following definition, which is also used in the proof of Lemma \ref{l:prdiff2}.

\smallskip
{\bf Definition of $a,s,\widehat{z^2}$ for $y\in \mathop{\fam0 ker}(2\rho_d\overline l)$.}
By Lemma \ref{l:res}.a there is a representative
$$a\in C_4(V)\quad\text{of}\quad Z\cap V\in H_4(V,\partial)\quad\text{such that}\quad\partial a=\partial\Delta-\xi P.$$
(Such a representative is obtained from a representative $a'\in C_4(V)$ of $Z\cap V\in H_4(V,\partial)$ such that $\partial a'=\partial\Delta-\xi P+\partial a''$ for some $a''\in C_4(\partial V)$ by the formula $a:=a'-a''$.)

Since $y\in\mathop{\fam0 ker}(2\rho_d\overline l)$, by \cite[Lemma 3.2.$\lambda$]{CSI}
there is a chain
$$s\in C_4(C_f\times0;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d)\quad\text{such that}\quad\partial s=2\xi P\times0.$$
Define
$$\widehat{z^2}:=[2a-2\Delta-2\xi P\times[0,\frac12]+s]\in H_4(W;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d).$$

\begin{proof}[Proof of Lemma \ref{l:prdiff2}]
We have
$$\rho_dz^2\cap W_-\overset{(1)}=2\rho_di_{V,W_-}(Z\cap V)=
[2a]_{W_-,\partial}=[2a-2\xi P\times[0,\frac12]+s]_{W_-,\partial}=\widehat{z^2}\cap W_-=j_W\widehat{z^2}\cap W_-,$$
where equality (1) follows by Lemma \ref{l:resa}.
Hence by the cohomology exact sequence of pair $(W,W_-)$ (cf.~diagram (*) in \cite[Proof of Lemma 4.8]{CSI})
$\rho_dz^2=j_W(\widehat{z^2}+n[t])$ for some $n\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$.
Thus property (a) holds.

Let us prove property (b).
We have $[t]^2=[S^2\times 0\times S^2]\capM{S^2\times D^3\times S^2}[S^2\times 1_3\times S^2]=0$.
Since the support of $\widehat{z^2}$ is in $W'\cup\partial C_f\times[0,\frac12]\cup C_f\times0$ and this space is the boundary of a connected component of $W-W'$, we have $(\widehat{z^2})^2=0$.
Also
$$[t]\capM{W}\widehat{z^2}=[t]\cap_{W_-} (\widehat{z^2}\cap W_-)=[t]\capM{W_-}[2a]_{W_-,\partial}=
2[t]\capM{\partial W_-}[\partial a]=2[t]\capM{t\times\partial\Delta}[\partial \Delta]=2.$$
Here the homology classes are taken in the space indicated under `$\cap$'
(so $[t]$ has different meaning in different parts of the formula),
and $\widehat{z^2}\cap W_-=[2a]_{W_-,\partial}$ is proved in the proof of (a).
\end{proof}

\begin{proof}[Proof of Lemma \ref{l:diff}.c]
Take any bundle isomorphism $\varphi:\partial C_0\to\partial C_1$ given by \cite[Lemma 2.5]{CSI}.
Take a closed oriented 3-submanifold $P\subset N$ realizing $y\in H_{f_0}=H_{f_1}$.
For $k=0,1$ construct maps $v_{jk}$, $j=0,1,2,3$, manifolds $V_k\subset C_k$, $\widehat{V_k}$, $W_k'$ and $W_k$,
chains $a_k,s_k$ and classes $Z_k,z_k,\widehat{z_k^2}$ as in \cite[Proof of Lemma 4.8]{CSI}
and above.
(So unlike in other parts of this paper, subscript 0 of a manifold does not mean deletion
of a codimension 0 ball from the manifold.)
Define
$$W := W_0\cup_{\varphi\times \mathop{\fam0 id} I:\partial C_0\times I\to\partial C_1\times I}W_1.$$
This $W$  should not be confused with what was previously denoted $W$ but now is denoted $W_0$ and $W_1$.
Same remark should be done for $z,Z,\widehat V$ constructed below.

Consider the following segment of the (`cohomological') Mayer-Vietoris sequence:
$$H_6(W,\partial)\overset{r_{W_0}\oplus r_{W_1}}\to H_6(W_0,\partial)\oplus H_6(W_1,\partial)\overset{r_0\oplus(-r_1)}
\to H_4(\partial C_0).$$
Here $r_k$ is the composition $H_6(W_k,\partial)\overset\partial\to H_5(\partial W_k)\overset{r_{\partial C_0}}\to H_4(\partial C_0)$.
We have
$$r_kZ_k= (\partial Z_k)\cap\partial C_0= Y_{f_k}\cap\partial C_0=\partial(Y_{f_k}\cap C_k)\overset{(4)}=
\partial A_k[N]\overset{(5)}=\partial A_{1-k}[N]\overset{(6)}= r_{1-k}Z_{1-k}\in H_4(\partial C_0).$$
Here

$\bullet$ equality (4) holds by descriptions of of joint Seifert classes \cite[Lemma 3.13.a]{CSI}.

$\bullet$ equality (5) holds by agreement of Seifert classes \cite[Lemma 3.5.a]{CSI}

$\bullet$ equality (6) holds analogously to the previous set of equalities.

Hence there exists $Z\in H_6(W,\partial)$ such that $Z\cap W_k=Z_k$.
Denote
$$\widehat V:=\widehat V_0\bigcup\limits_{\varphi:\nu_0^{-1}P\to\nu_1^{-1}P}\widehat V_1\subset W'\quad\text{and}\quad
z:=Z+[\widehat V]\in H_6(W,\partial).$$
Clearly,  $z\cap W_k=z_k$.
\footnote{Note that $(W,z)$ is not assumed to be spin bordism of anything and possibly $\rho_d\partial z^2\ne0$.}

Take $\widehat{z_k^2}\in H_4(W;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d)$ given by Lemma \ref{l:prdiff2}.
Then by Lemmas \ref{l:res}.b and \ref{l:prdiff2}
$$\overline{z_k^2}\capM{W} p_W^*=4m_k=\widehat{z_k^2}\capM{W} p_W^*\quad\text{and}\quad
\overline{z_k^2}\capM{W} z_k^2=\overline{z_k^2}\capM{W}\overline{z_k^2}=4n_k=
2\widehat{z_k^2}\capM{W} \overline{z_k^2}=2\widehat{z_k^2}\capM{W} z_k^2.$$
Hence
$$\eta(f_k,y)=\rho_{\widehat d}(\widehat{z_k^2}\capM{W_k}(2z_k^2-p_{W_k}^*))=
\rho_{\widehat d}(\widehat{z_k^2}\capM{W}(2z^2-p_W^*)).$$
Take a weakly unlinked section $\xi_0:N_0\to\partial C_0$ of $f_0$.
By \cite[Lemma 3.4]{CSI}
$\xi_1:=\varphi\xi_0$ is an unlinked section of $f_1$.
Hence
$$\partial a_1-\partial\Delta_1=-\xi_1P=-\xi_0P=\partial a_0-\partial\Delta_0\quad\text{and}\quad \partial s_1=2\xi_1P=2\xi_0P=\partial s_0.$$
Identify $M_\varphi$ and its subsets with $M_\varphi\times0\subset\partial W$ and with corresponding subsets.
Denote
$$\widehat a:=[\Delta_0-a_0+a_1-\Delta_1]\in H_4(\widehat V;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d)
\quad\text{and}\quad s:=[s_0-s_1]\in H_4(M_\varphi;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d).$$
Then by definition of $\widehat{z_k^2}$
$$\widehat{z_0^2}-\widehat{z_1^2}=i_\varphi s-2i\widehat a,
\quad\text{where}\quad i_\varphi:=i_{M_\varphi,W}\quad\text{and}\quad  i:=i_{\widehat V,W}.$$
We have $i_\varphi s\capM{W} p_W^* =s\capM{M_\varphi} p_{M_\varphi}^*=0$.

Since
$$(z\cap M_\varphi)\capM{M_\varphi}S^2_{f_0}=(\partial z_0\cap C_0)\capM{C_0}S^2_{f_0}=Y_{f_0}\capM{C_0}S^2_{f_0}=1,$$
$z\cap M_\varphi$ is a joint Seifert class for $\varphi$.
Then
$$i_\varphi s\capM{W} z^2=(s\cap\partial C_0)\capM{\partial C_0}(z^2\cap\partial C_0)\overset{(2)}=
2\xi_0 y\capM{\partial C_0}\nu_0^! \beta= 2\beta\capM{N} y,$$
where

$\bullet$ $\beta\in H_1(N;{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d)$ is a lifting of $\beta(f_0,f_1)$;

$\bullet$ equality (2) follows because $s\cap\partial C_0=2[\xi_0 P]=2\xi_0 y$ and because
$z^2\cap\partial C_0=(z\cap M_\varphi)^2\cap\partial C_0=\nu_0^! \beta$ by definition of $\beta(f_0,f_1)$.

We have
$$z^2\capM{W} i\widehat a \overset{(1)}=
(Z+[\widehat V])^2\capM{W} i\widehat a\overset{(2)}=
Z^2\capM{W} i\widehat a+2Z\capM{W}[\widehat V]\capM{W} i\widehat a\overset{(3)}=$$
$$=(Z\cap\widehat V)^2_{\widehat V}\capM{\widehat V}\widehat a+2i(Z\cap\widehat V)\capM{W} i\widehat a\overset{(4)}=
(\widehat a)^3_{\widehat V}+2(i\widehat a)^2\overset{(5)}=(\widehat a)^3_{\widehat V},$$
where

$\bullet$ equality (1) follows by definition of $z$;

$\bullet$ equalities (2) and (5) follow because $\widehat V\subset W'$, so $[\widehat V]^2=0$ and $(i\widehat a)^2=0$;

$\bullet$ equality (3) is obvious;

$\bullet$ equality (4) follows because $Z\cap\widehat V=\widehat a$ by definition of $a_0,a_1,\widehat a$.

Therefore
$i\widehat a\capM{W}(2z^2-p_W^*) = 2(\widehat a)^3_{\widehat V}-\widehat a\cap_{\widehat V}p_{\widehat V}^* \underset{12}\equiv0$ by \cite[Theorem 5]{Wa66}.

Now the Lemma follows because
$$(\widehat{z_0^2}-\widehat{z_1^2})\capM{W}(2z^2-p_W^*)=
2i_\varphi s\capM{W} z^2-i_\varphi s\capM{W} p_W^*-2i\widehat a\capM{W}(2z^2-p_W^*)
\underset{24}\equiv 4\beta\capM{N} y. $$
\end{proof}

\comment

to eta-invariant

We have
$$H_s(\partial C_1\times I,U_1)\overset\mathop{\fam0 ex}\cong H_s(\Sigma_1,\partial)\cong H^{7-s}(\Sigma_1)=0\quad\text{for}\quad s=2,3,4,6.$$
Hence from the exact sequence of pair $(\partial C_1\times I,U_1)$ we obtain that the inclusion
$H_s(U_1)\to H_s(\partial C_1\times I)$ is an isomorphism for $s=3,4$, a monomorphism for $s=5$ and
an epimorphism for $s=6$.
Looking at the inclusion-induced mapping of the exact sequences of pairs
$(M_\varphi,C_0)$ and $(V,C_0\times I)$, by the 5-lemma we obtain that the inclusion $M_\varphi\to V$
induces an isomorphism in $H_s(\cdot)$ for $s=3,4,5,6$ (for $s=5$ latter using 5-lemma in an improved form).
How to arrive to $r_{M_\varphi}\partial$?

{\it Attempt of the proof of the second assertion.}
Since $H_4(S^2\tilde\times S^5)=0$, analogously to \cite[Proof of the Independence Lemma]{Sk08'} and
Lemma \ref{l:cobordbeta} we obtain that for each $x\in H_3$

{\it the triple $(M_{\varphi'},Y_4',Y_3')$ is cobordant to $(M_\varphi,Y_4,Y_3)\sqcup(S^2\tilde\times S^5, A,0)$ for some joint Seifert classs $Y_s\in H_{s+1}(M_\varphi)$ and $Y_s'\in H_{s+1}(M_{\varphi'})$, where $Y_4=Y$ and $Y_3$, $Y_3'$ are for $x$.}

Since $Y_4=Y$ is a $d$-class, we have $\rho_d((Y_4')^2\cap Y_3')=\rho_d(Y_4^2\cap Y_3+A^2\cap0)=0$
for each $x\in H_3$ and some $Y_3'\in H_4(M_\varphi,C_1)$ for $x$.
Hence??? $\rho_dj_{M_\varphi,C_0}(Y_4')^2=0\in H_3(M_\varphi,C_0)$.
Therefore $(Y_4')^2\in \mathop{\fam0 im} i_{M_\varphi,C_0}$.
Although $\beta(f_0,f_1)=0$, analogously to the well-definition of $\beta$ (Lemma \ref{l:webeta})
we do not obtain that $\rho_d(Y_4')^2=0$.
Now, we obtain that for each $x\in H_3$ there is a ? $Y_3'$ for $x$ such that $\rho_d((Y_4')^2\cap Y_3')=0$. But this does not imply that $\rho_d(Y_4')^2=0$!
Unlike the proof of the transitivity!

 $r_{M_\varphi}\partial:H_s(V,\partial)\to H_{s-1}(M_\varphi)$ is an isomorphism for $s=4,6$.
 can also be proved by mapping Mayer-Vietoris sequences induced by $r_{M_\varphi}\partial$.)

Consider (the Poincar\'e dual of) the Mayer-Vietoris sequence for $V$:
$$H_6(U_1,\partial)\overset{i_{V,U_1}}\to H_6(V,\partial)\overset{r_{V,C_0\times I}\oplus r_{V,C_1\times I}}
\to H_6(C_0\times I,\partial)\oplus H_6(C_1\times I,\partial)
\overset{-r_{\partial C_1\times I,U_1}\overline\alpha\partial\oplus r_{\partial C_1\times I,U_1}\partial}\to H_5(U_1,\partial).$$

By the agreement of Seifert classs (Lemma \ref{l:Agr}.a) $\varphi\partial A_0=\partial A_1$ and
$\varphi'\partial A_0=\partial A_1$.
Hence there is a class
$$\overline Y\in H_6(V,\partial)\quad\text{such that}\quad
\overline Y\cap (C_k\times I)=A_k[N]\times I\in H_6(C_k\times I,\partial)\quad\text{for each}\quad k=0,1.$$
Then by  \ref{l:desei}.a $Y:=\partial\overline Y\cap M_\varphi$ and $Y':=\partial\overline Y\cap M_{\varphi'}$ are joint Seifert classs.
The choice of $\overline Y$ is in $H_6(U_1,\partial)\cong H^1(U_1)$.
We have
$$H^s(\partial C_1\times I,U_1)\overset\mathop{\fam0 ex}\cong H^s(\Sigma_1,\partial)\cong H_{7-s}(\Sigma_1)=0\quad\text{for}\quad s=1,2.$$
Hence from the exact sequence of pair $(\partial C_1\times I,U_1)$ we obtain that the restriction
$H^1(\partial C_1\times I)\to H^1(U_1)$ is an isomorphism.
In dual form this means that $r_{\partial C_1\times I,U_1}:H_6(\partial C_1\times I,\partial)\to H_6(U_1,\partial)$ is an isomorphism.

For each $y\in H_3$ let $\overline Y_y:=\overline Y+i_{V,U_1}(\nu^!y\times I)\cap U_1$.
Analogously to Lemma \ref{l:desei}.c
$\overline Y_y^2-\overline Y^2=2i_{V,U_1}(\nu^!y\overline{\lambda(f_0)}(y)\times I)\cap U_1$...

{\bf Remarks.}
 An embedding $f:N\to S^7$ is {\it $\eta$-null} if
$$\eta(\mathop{\fam0 id}\phantom{}_{\partial C_f},Y)=0\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_{(d(u),24)}\quad\text{for each  $d(u)$-class}\quad Y\in H_5(M_f).$$

Lemma \label{l:merc}
(a) If $K:=\mathop{\fam0 ker}(2\rho_d\overline l)\subset H_3$ is a direct summand, then there is an element $b\in\mathop{\fam0 coker}(2\rho_d\overline l)$ such that $\eta_{u,l,b}=0$.

(b) If $\eta_{u,l,b}=\eta_{u,l,b'}=0$, then $\Delta(d,2\overline l,b-b')$ is divisible by 6.

Proof of (b). Take any $y\in \mathop{\fam0 ker}(2\rho_d\overline l)$.
Since $0=\eta_{u,l,b}(y)-\eta_{u,l,b'}(y)=4\rho_{(d,24)}(b-b')\cap y$, the number $4(b-b')\cap y$ is divisible by $(d,24)$.
So $(b-b')\cap y$ is divisible by 6.
Hence by the following Lemma \ref{l:unimzd} $\Delta(d(u),2\overline l,b-b')$ is divisible by 6.

{\it Old statement of Lemma \ref{l:merc}.a.}

(a) If $(\varkappa\times\lambda)^{-1}(0,l)$ is non-empty, then it contains an $\eta$-null embedding.
\footnote{Trivially correct for $2\rho_{d(u)}\overline l=0$.
Presumably correct for $u$ not divisible by 8 or 3, but false in general.
Indeed, under the assumptions of Lemma \ref{l:unimzd}
for $d\ne0$ and a homomorphism $c:\mathop{\fam0 ker}(\rho_dm)\to{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$ there could be no homomorphisms $\overline c:V\to Z$ such that $c=\rho_d\overline c|_{\mathop{\fam0 ker}(\rho_dm)}$.
Indeed, let $V={\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$, \ `$m=\mathop{\fam0 id}$', i.e.~$m(x)y:=xy$, \ $d=2$.
Then $\mathop{\fam0 ker}(\rho_2m)=2{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$.
Define a homomorphism $c:2{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}\to{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_2$ by $c(2x)=\rho_2x$.
Then $c$ does not extend to ${\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$.
If $c(y)=\eta(f,y)$ for some embedding $f:N\to S^7$, then $(\varkappa\times\lambda)^{-1}(\varkappa\times\lambda)(f)$ does not contain an $\eta$-null embedding.}

(b) If $f,g\in(\varkappa\times\lambda)^{-1}(u,l)$ are $\eta$-null embeddings,
then $\Delta(d(u),2\overline l,\beta(f,g))$
is divisible by 6.

\begin{proof}[Attempt of proof of Lemma \ref{l:merc}.a in the old statement for general $d$]
Take any $f\in(\varkappa\times\lambda)^{-1}(u,l)$.
In this proof let $K:=\mathop{\fam0 ker}(2\rho_d\overline l)$.
Since $H_3$ is free, $K$ is free.
Let $y_1,\dots,y_s$ be free generators of $K$.
By Lemma \ref{l:diff}.a for each $n=1,2,\dots,s$ there is a lift $4\widetilde b_n\in 4{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$ of $\eta(f,y_n)$.
A map $\widetilde  b:K\to {\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}_d$ is well-defined by
$\widetilde  b(\sum_n\lambda_ny_n):=\sum_n\lambda_n\widetilde  b_n$ for each $\lambda_1,\dots,\lambda_s\in{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\C{{\mathbb C}} \def\Ha{{\mathbb H}$.
By Lemma \ref{l:diff}.b $\eta(f,y)=4\rho_{(d,24)}\widetilde  b(y)$ for each $y\in K$.
The subgroup $K\subset H_3$ is a direct summand {\bf only for $d=0$}.
Hence by Poincar\'e duality there is a class $b\in H_1$ such that $\widetilde  b(y)=\rho_d(b\cap y)$ for each $y\in K$.
By the $\beta$-transitivity and the parametric additivity (Lemmas \ref{l:Dif} and \ref{l:Pad})
there is an element $f'\in(\varkappa\times\lambda)^{-1}(u,l)$ such that $\beta(f,f')=[b]$.
Then  by Lemma \ref{l:diff}.c $\eta(f',y)=\eta(f,y)-4\rho_{24}(b\cap y)=0$ for each $y\in K$.
Hence by Lemma \ref{l:simpleY0} $f'$ is $\eta$-null.
\end{proof}

\begin{proof}[Proof of Lemma \ref{l:merc}.b in the old statement]
Since $f$ and $g$ are $\eta$-null, by Lemma \ref{l:diff}.c for each
$y\in \mathop{\fam0 ker}(2\rho_d\overline l)$ the number $4\beta(f,g)\cap y$ is divisible by $(d,24)$.
So $\beta(f,g)\cap y$ is divisible by 6.
Hence by the following Lemma \ref{l:unimzd} $\Delta(d(u),2\overline l,\beta(f,g))$ is divisible by 6.
\end{proof}

\endcomment

redpajama_set_name: RedPajamaArXiv


This website is a bit of a work in progress. A couple of these projects have more details inside. More soon.
My name is Stuart Geddes. This is some of my work.
This is also the work of my collaborators: authors, editors, publishers, printers and other designers. I mostly design books, but I sometimes work on other kinds of projects, like magazines and journals, exhibitions and websites.
I'm an Industry Fellow at RMIT University, as well as being a PhD candidate. My research interests in that context converge around the form of the book, through collaborative practice, emerging histories, and unconventional economies.
One of my ongoing projects is making motorcycle magazine Head Full oƒ Snakes with my friend Luke. And I share a studio space with Brad, Žiga, Karina and Tom.
Before all this I ran Chase & Galley, I used to make Is Not Magazine, design Monument, and be on the board of West Space and AGDA. Before all that, I was part of Studio Anybody. Along the way I've taught quite a lot at RMIT and Monash Universities, and frequently given talks and run workshops for conferences, festivals, galleries, schools and other organisations.


Description : Download editabletemplates.com's cost-effective Spiral Lollipop Microsoft Word Template Design now. This Spiral Lollipop editable Word template is royalty free and easy to use. Be effective with your business documentation by simply putting your content in our Spiral Lollipop Word document template, which is very economical yet does not compromise with quality.


text:
Владимир Петрович Жарков (6 января 1924 — 1 октября 1943) — красноармеец Рабоче-крестьянской Красной Армии, участник Великой Отечественной войны, Герой Советского Союза (1943).

Биография

Владимир Жарков родился в 1924 году в деревне Шевертни (ныне — Гусь-Хрустальный район Владимирской области). После окончания школы работал в колхозе. С августа 1942 года — на фронтах Великой Отечественной войны. Принимал участие в боях на Сталинградском и Южном фронтах. К октябрю 1943 года гвардии красноармеец Владимир Жарков был пулемётчиком 2-го стрелкового батальона 262-го гвардейского стрелкового полка 87-й гвардейской стрелковой дивизии 13-го гвардейского стрелкового корпуса 2-й гвардейской армии Южного фронта. Отличился во время Мелитопольской операции.

В боях к северу от Мелитополя Жарков уничтожил ряд вражеских огневых точек и неоднократно срывал выдвижение вражеских резервов, чем способствовал успешному продвижению вперёд всего подразделения. 1 октября 1943 года у населённого пункта Нейдорф (ныне — хутор Ровный в черте села Коханое Токмакского района Запорожской области Украины), когда немецкие войска предприняли контратаку, Жарков пулемётным огнём уничтожил около 60 вражеских солдат и офицеров. В том бою он был тяжело ранен, но продолжал сражаться, пока не погиб. Похоронен в братской могиле в селе Коханое.

Указом Президиума Верховного Совета СССР от 1 ноября 1943 года за «образцовое выполнение боевых заданий командования по прорыву укрепленной полосы немцев и освобождению города Мелитополь и проявленные при этом отвагу и геройство» гвардии красноармеец Владимир Жарков посмертно был удостоен высокого звания Героя Советского Союза. Также посмертно был награждён орденом Ленина.

Примечания

Литература

 Москвитин Я.П., Нагорный А.Ф. Твои Герои, земля Владимирская. - Владимир, 2000.

Участники освобождения Мелитополя
Погибшие в боях Великой Отечественной войны
Похороненные в Токмакском районе

redpajama_set_name: RedPajamaWikipedia


The Sports Boxing – Crawford was hovering around with aim and was scarcely gotten in the opening rounds yet the two glimmer knockdowns in the fifth set the example by online sports. The principal shot after the chime, a straight right, saw Pistol bumble and regardless of the possibility that the elusive canvas accomplished more harm to his parity then the punch, his knee touched the floor leaving ref Tony Weeks with no decision yet to call a check.
The second knockdown, however, demonstrated the force that Crawford has nearby his smoothness, with an overwhelming left to the body sending the Ukrainian staggering back and the accompanying right driving him to touch down once more, this time with his glove.
Crawford was in finished control to win the unification battle.
The two numbers implied that Postol was compelled to press in the second 50% of the battle yet his upright position and one-dimensional development left him consistently got by countering shots, with Crawford likewise eager to demonstrate his hostile ability to an ever increasing extent.
Another substantial left created his rival to lose his equalization again in the eighth, a fast whirlwind doing likewise in the ninth and his southpaw punch and catch up left arriving freely.
The eleventh round summed up the battle, with Pistol cautioned and afterward deducted a point for hitting round the back of the head on the uncommon event the pair were in a secure, while Crawford essentially grinned, cooled off then returned with another tasteful mix.
Be that as it may, knowing the malaysia sports boxing scene was watching and quick to affirm himself as the best in the super-lightweight division, the American was cheerful to remain nearby again in the last round, this time, a peach of a short, sharp right denying Post no less than a positive completion to his bad dream.


text:
Q: How can I set Extra Data into Access Token in Authorization code flow I am implementing a OAuth2 provider using DotNetOpenAuth.
Using implicit flow, I was capable of passing extra data to function CreateAccessToken (IAuthorizationServerHost), inside the param (IAccessTokenRequest accessTokenRequestMessage) and assign that to the token:
accessToken.ExtraData.Add("DeptID", accessTokenRequestMessage.ExtraData["DeptID"].ToString());

I accomplished it adding the info inside EndUserAuthorizationRequest in Auhtorize endpoint.
The issue is that using authorization code flow, and doing the same thing, the param (IAccessTokenRequest accessTokenRequestMessage) always has an empty collection of ExtraData.
I suppose that it is because using this flow there is a second call to Token endpoint, and the information is lost.
How could I pass this information to CreateAccessToken function?
Thanks in advance.

A: The only way I achieved to pass information to the token creation was using the User property:
var response = this.AuthServer.PrepareApproveAuthorizationRequest(AuthorizationRequest, userInfoString);

I have passed a encrypted string of JSON serialization of the information needed. Then, from CreateAccessToken function I've retrieved the value using Reflection over IAccessTokenRequest parameter.
It is not a solution as elegant as I wish, but at least it worked!

redpajama_set_name: RedPajamaStackExchange


text:
About FEDSM
search filter All ContentAll ProceedingsMultidisciplinary FEDSM
ASME 2017 Fluids Engineering Division Summer Meeting
July 30–August 3, 2017
Waikoloa, Hawaii, USA
Fluids Engineering Division
Volume 2, Fora: Cavitation and Multiphase Flow; Advances in Fluids Engineering Education
Development and Commissioning of a Supersonic Blow Down Wind Tunnel for Educational Purposes
Maximilian Passmann
Muenster University of Applied Sciences, Steinfurt, Germany
Stefan aus der Wiesche
Paper No: FEDSM2017-69196, V002T17A002; 10 pages
https://doi.org/10.1115/FEDSM2017-69196
Published Online: October 24, 2017
Passmann, Maximilian, and aus der Wiesche, Stefan. "Development and Commissioning of a Supersonic Blow Down Wind Tunnel for Educational Purposes." Proceedings of the ASME 2017 Fluids Engineering Division Summer Meeting. Volume 2, Fora: Cavitation and Multiphase Flow; Advances in Fluids Engineering Education. Waikoloa, Hawaii, USA. July 30–August 3, 2017. V002T17A002. ASME. https://doi.org/10.1115/FEDSM2017-69196
A cost-effective test rig is presented that allows for the experimental investigation of supersonic flows for educational purposes. The individual units for the test rig were designed and built by students as part of their degrees. The test rig allows for operating times up to 10 seconds and features a nozzle test section, that can house different test objects. The divergent part of the de Laval nozzle geometry is designed using the method of characteristics for planar two-dimensional supersonic flow. State of the art 3D printing technology has been utilized to manufacture the nozzle geometry. Both optical and pneumatic measurement techniques have been adopted for the current setup. A z-type schlieren setup with two parabolic mirrors is used to perform flow visualization. The entire run can be recorded with a digital high speed camera. Stagnation pressure and temperature are measured in the pressure reservoir. Measurements are used to demonstrate basic thermodynamic effects such as the depressurization of gas-filled pressure vessels. Schlieren photography is used to graphically derive the Mach number and some aspects of Mach waves, oblique shock waves, and expansion waves are discussed. Finally, some effects of surface roughness on the flow field are addressed in this paper. Initial tests with the untreated nozzle geometry led to a fine pattern of very weak oblique shock waves in the supersonic part of the nozzle, that were caused by the finite layer thickness of the printer.
Forum on Advances in Fluids Engineering Education
Additive manufacturing, Flow (Dynamics), Flow visualization, Geometry, Mach number, Mirrors, Nozzles, Pressure, Pressure vessels, Reservoirs, Schlieren methods, Shock waves, Students, Supersonic flow, Surface roughness, Temperature, Waves, Wind tunnels
A Study of Radial Underexpanded Jet by Shadow and Schlieren Photography
FEDSM2007
Experimental Study of Mixed Compression Air-Intake for Hypersonic Airbreathing Engines
Design of a Test Rig for Organic Vapours
Supersonic Flow at Micro-Tube Outlet
Improvement of the Performance of a Supersonic Nozzle by Riblets
J. Fluids Eng (September, 2000)
Pressure Losses in Combining Subsonic Flows Through Branched Ducts
J. Turbomach (January, 1992)
Design of the Dense Gas Flexible Asymmetric Shock Tube
J. Fluids Eng (March, 2008)
A Simple Carburetor
Case Studies in Fluid Mechanics with Sensitivities to Governing Variables
Hydrodynamic Stabilization of Supercavitating Underwater Bodies
Proceedings of the 10th International Symposium on Cavitation (CAV2018)
Development of Nuclear Boiler and Pressure Vessels in Taiwan
Companion Guide to the ASME Boiler and Pressure Vessel Code, Volume 3, Third Edition

redpajama_set_name: RedPajamaCommonCrawl


Leep The Bridge is a club night for people with learning disabilities. We have one club night every month. It is for people who live in Leeds. We have our club nights at PRYZM nightclub in the centre of Leeds.
We received funding from Ideas that Change Lives, so we can create monthly club nights for adults with learning disabilities. These club nights are held at PRYZM and other locations around Leeds.
Our club nights give you the opportunity to catch up with friends, meet new people, have a good time enjoying yourself and dancing the night away. These are always themed nights.
Here is the poster for the next club night on Thursday 18th April 2019 and the dates for the rest of 2019.


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text:
Home » Congress News » What's the timeline for…
What's the timeline for a Kavanaugh vote in the Senate?
Mitchell Miller | @mmillerwtop
October 2, 2018, 4:59 PM
Even if the FBI quickly wraps up, the Senate's procedural process would need to start up within the next couple of days for a final vote to come by the end of the week.
WASHINGTON — The FBI is investigating, lawmakers are sniping and the nation is watching. So when will a final vote be taken in the Senate on Supreme Court nominee Brett Kavanaugh?
'Hit the ground running' — FBI has time for a thorough Kavanaugh review, ex-agent says
What we know and don't know about the Senate impeachment trial
On Tuesday, Majority Leader Mitch McConnell (R-Ky.) renewed his pledge to vote this week. But exactly what day that would be is still unclear.
"After the FBI shares what they've found, senators will have the opportunity to vote," McConnell said. "They'll have the opportunity to vote 'no' on the politics of personal destruction. They'll have the opportunity to vote 'yes' on this fine nominee."
The FBI investigation into allegations of sexual misconduct against Kavanaugh — all of which he's denied — is the crucial factor in determining when a vote could begin. At the request of Sen. Jeff Flake (R-Ariz.) and other lawmakers, the Senate last Friday agreed the FBI could investigate for up to a week, meaning its deadline is this Friday. But the FBI could complete its interviews at any time.
Republicans have accused Democrats of trying to drag out the confirmation process, with McConnell repeatedly charging they're trying to "move the goalposts."
Minority Leader Chuck Schumer (D-N.Y.) denied that in a Senate floor speech Tuesday.
"No Democrat has called for taking more than a week. We are not moving the goalposts back," Schumer said.
But just a few hours later, the ranking Democrat on the Senate Judiciary Committee, Sen. Dianne Feinstein (D-Calif.), told reporters that a Friday vote on Kavanaugh would be "too soon," adding that it wouldn't give lawmakers enough time to review the FBI probe.
The chairman of the committee, Sen. Chuck Grassley (R-Iowa), said Tuesday that he expected the FBI report to be released before Friday, but that he wasn't sure exactly when.
Schumer has made a request to McConnell that all senators be briefed on the FBI's findings at least 24 hours before a Senate vote to end debate on the nomination.
McConnell doesn't want the wait for a vote to go on much longer. But he also can't afford to be seen as rushing, in the wake of Flake's request for the delay, which is supported by other key Republicans, Sen. Lisa Murkowski (R-Alaska) and Sen. Susan Collins (R-Maine).
Flake has said he will vote for Kavanaugh, while Murkowski and Collins have not announced what they'll do. If all Democratic senators vote against Kavanaugh, Republicans — who have a 51-49 edge in the Senate — can only afford to lose one vote to get the judge's confirmation passed.
The Senate has already let several earlier deadlines go by. Everything changed after Christine Blasey Ford agreed to testify at last week's highly charged hearing on Capitol Hill.
And while the nomination process is moving forward, it remains fluid. There's also Senate procedure, which takes time. Over a period of several days, McConnell would need to end debate on the nomination; provide a one-day waiting period; and then the Senate would need to vote on a cloture motion. That would open things up for the final Senate debate on the nomination, which can last up to 30 hours.
Kavanaugh first testified for his confirmation hearing before lawmakers on the committee nearly a month ago – Sept. 4.
Vehicles DC-area original owners drive the longest
Congress News Government News Supreme Court News
Brett Kavanaugh mitchell miller


It's a paradox. Scientifically, black is empty of color, and white, the color of light, contains all the colors. But an artist wishing to paint the color black selects all of the darkest colors on his/her palette and mixes them together. It's a delicate balance to keep one of the original colors from dominating, but done right, it creates black. So there are two kinds of black–the kind that is devoid of color, and the kind that contains all the colors.
Lately, I have been talking about painting all the colors I can find in Caucasian skin. "White" skin. More accurately, light skin. Today I have two new paintings of a "black" model. More accurately, a really dark-skinned model. Not my first really dark skin, by any means, but my first while consciously searching for more color in skin tones. I honestly don't know, as I put this blog together, whether being color-conscious has made any difference.
From the looks of it, I used a lot of burnt sienna and ochre.
Looks like I started with burnt umber. Sabrin's arm photographed grayer than the actual painting, so the color here is not quite as awful as it looks. I used a lot of purple here.
I was in such a hurry–the model was very late to arrive, that my choice of colors here must have been pure instinct, based on what I saw. Let it be acknowledged, however, that the model's skin was quite a dark brown, not caramel brown.
Then, just a few weeks ago, I posted this painting of Grace in the yellow chair, which I believe is my first live painting of her. Apparently, I found some red in her.
All of that was done without much thought. Some thought, just not as much as I have been wallowing in lately. Gosh, I hope that's a sign I am becoming a better artist–more time to wallow in thought.
I'm skipping over the next two–one because of the hot pink drape that I hated, and the other because I hated the whole thing (although I have not yet destroyed it). If you feel compelled to check it out, the hot pink drape was two weeks ago, and all-bad one here.
I hope you can tell which was the earlier, because I tackled the second one with even more resolve to find colors other than brown. One problem I have with the darker, brown base is that the color tends to gray down if you just add white, like that arm in Sabrin. Lighter colors seem easier to keep pure. No, that's just an excuse. The truth is, I couldn't see any colors that were not variations of brown. Mostly warm browns. My model said other people have claimed that they see blue in her skin. Not me. Maybe some navy blue in the shadows. Maybe some hint of sky blue in the highlights (as I found on the Girl in the Red Headdress). Maybe purple in the shadows. But no aquas, no greens.
I am loving the brown sofa painting, regardless of color issue–the pose and composition are very pleasing. Almost lost the foot, but managed to rescue it by foreshortening the lower limb. Also, I discovered, with my new eye (cataract removed two weeks ago), that the blue drape reflected the brown of Grace's skin, which was kind of startling. Note that the brown leather sofa is a match to the brown leather recliner. If we keep this up, you are going to become familiar with every piece of furniture in Adrienne and Heather's studio.
Grace is back on deck this Tuesday. Maybe I should start by drawing her figure in green or aqua, instead of burnt sienna, just to see what effect it has. Not kidding.
This entry was posted in art, art artist painter, artist, Nudes, painter, Paintings, Portraits and tagged nudes. Bookmark the permalink.
Oh, Aline, On the Brown Leather Sofa is the best yet. Before I came to your words about loving this one yourself, I had already said that in my mind. It is great!
No paradox – scientific or otherwise – when you understand the PHYSICS of color. Every surface either absorbs or reflects light waves. The color we SEE is the color (or combination of colors) that are reflected back (NOT absorbed). A surface that appears black absorbs all light waves/energy – that is (one reason) why black cloth feels hot. A surface that appears white reflects ALL light waves, and they remain unseparated so that we can't distinguish any color. So – technically, black DOES contain ALL colors (because they are absorbed) and we will see the blues or purples or reds as a result of surface irregularities that might let small amounts of certain wavelengths escape. Same principle applies to very pale shades of colors being absorbed in what is white. The entire industries that are responsible for producing pigments, dyes, paints, etc depend on understanding and applying the PHYSICS … and those rules are very different from what we generally learn about color mixing (principles of which are based on pre-made product). It CAN get confusing, but I think it's easier to remember (if you really need to) when you think of black or white fabric in summer sun and which one gets hot!
I still like the Girl in the Red Turban!
You always take the mystery out of things.
You are producing some lovely work Aline. I really prefer the first of the latest two. I think it is because the brushwork is less fussy and overall it looks more assured.