“adjoint” =?= “inverse of composite endofunctor is uniform bi-composition”

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Understanding adjoints has always been (and continues to be) a bit of a struggle for me.

Today I stumbled upon a property of adjoint functors which seemed extremely intuitive to me. I was wondering why this property isn't mentioned more often in introductory category theory literature, and whether or not it completely characterizes adjunctions.

If two functors $F:C\to D$ and $U:D\to C$ are adjoint $F\dashv U$, then for every $f:F(Y)\to X$ in $D$ there exists an $\hat f:Y\to U(X)$ in $C$ such that

$$U(f)\circ \eta_Y = \hat f$$ $$\epsilon_X\circ F(\hat f)=f$$

If we substitute the top equation into the bottom, we get

$$\epsilon_X\circ F(U(f)\circ \eta_Y)=f$$

and by functoriality we get

$$\epsilon_X\circ F(U(f))\circ F(\eta_Y)=f$$ $$\epsilon_X\circ (F\circ U)(f)\circ F(\eta_Y)=f$$

What the last equation says is that we can recover any morphism $f$ from the action of the "round trip endofunctor" $F\circ U$ on it by pre-composing with $\epsilon_X$ and post-composing with $F(\eta_Y)$. These two morphisms are determined only by the domain and codomain of $f$ -- we only needed to know $X$ and $F(Y)$ in order to pick the two morphisms. We would have picked the same two morphisms for some $g\neq f$ as long as $g:F(Y)\to X$.

So, I believe it is correct to say that "if the domain of a morphism is within the range of a functor which has a right adjoint, then it can be recovered from the action of the composite endofunctor on it by pre-composition with some morphism and post-composition with some other morphism, where the choice of these two morphisms is completely determined by the domain and codomain of the original morphism". There is, of course, an equivalent statement for morphisms with a codomain in the range of a functor with a left adjoint.

So, my three questions are: (1) is this correct, (2) if so, why isn't it used to explain adjunctions to beginners (I certainly would have caught on quicker!) and (3) does the condition completely characterize adjoint functors?

Thanks,

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I don't understand why you say that "if we want to know the action of FU on f, we just pre- and post- compose with $\epsilon$ and $F(\eta)$". Doesn't your formula (which is true) really say that "f can be recovered from FU(f), by pre- and post-composing ..."? – Charles Rezk Mar 23 2010 at 20:23
Wow, thank you, yes that was a glaring error. I have modified the question and its title to correct this. Thank you! – Adam Mar 23 2010 at 20:41
The condition you wrote + its dual + the assertion that $\eta:I_C\to UF$ and $\varepsilon:FU\to I_D$ are natural totally characterize the adjunction, because an adjunction is totally characterized by the triangular identities $U\varepsilon\circ\eta U=I_U$, $\varepsilon F\circ F\eta=I_F$, see part (v) of Theorem IV.2 on page 83 of Mac Lane. To get the triangular identities from yours, just substitute $f=1_{FY}$ (so that $X=FY$), and similarly in the dual. BTW, the derivation of the triangular identities is very similar to yours, see p. 82 in Mac Lane. – unknown (google) Mar 23 2010 at 22:55
This might sound dumb, but I never understood the $F\eta$ notation. If $F$ is a functor and $\eta$ is a morphism of functors, what does $F\eta$ mean? It can't be composition because they aren't both morphisms in the same category... and viewing $\eta$ as an object-indexed family of morphisms doesn't seem to work out either because $F$ isn't an object. Does this mean reindexing? – Adam Mar 24 2010 at 0:21
$F\eta$ is the natural transformation whose component on an object $x$ is the map $F(\eta_x)$. – Reid Barton Mar 24 2010 at 2:12
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1 Answer

(1) Yes. (2) Well, it doesn't give me any additional intuition. You didn't say why it helps you understand, so I can't judge what the advantage of it might be. I think this is really just a complicated way of giving the "bijection of hom-sets" condition.

(3) No, you need something more. For instance, let $r:B\to A$ be a surjection with section $s$, let $C$ have two objects $x$ and $y$ with $C(x,y)=B$, $C(y,x)=\emptyset$, and $C(x,x)=C(y,y)=1$ (only identities), let $D$ be similar using $A$ instead, and let $F:C\to D$ and $U:D\to C$ be the identity on objects and with action on arrows given by $r$ and $s$ respectively. Pick $\varepsilon$ and $\eta$ to be identities. Then every morphism in $D$ can be recovered, as you describe, but the components of $\eta$ are not natural, and the dual condition fails.

The "unknown (google)" comment above explained why if you additionally require the dual condition, plus naturality of $\eta$ and $\varepsilon$, then you do get an adjunction. (Although it's not clear to me from the condition you stated whether you wanted to require the morphism playing the role of $F(\eta)$ to actually be $F$ of something, which is also necessary for this argument to work.)

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I think this is really just a complicated way of giving the "bijection of hom-sets" condition. -- Except that it never mentions sets. That's the advantage. The whole notion of hom-**sets** (rather than hom-objects) still seems weird and unusual to me, and I'm always afraid that some of my deep-seated intuitions about sets are going to limit my thinking. So I try to block the concept of hom-set out of my mind. The definition above is something that would make sense to a beginner long before they're ready to learn about enrichment. – Adam Oct 11 2010 at 9:16

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But it still mentions elements of hom-sets, i.e. single morphisms, and therefore does not carry over to the enriched world. – Mike Shulman Oct 11 2010 at 17:17
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