# Groups in Category Theory

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• November 21st 2011, 02:50 PM
albi
Groups in Category Theory
There are some ways to introduce groups in Category Theory.

1. Given object $A$ and morphisms $\mu: A \times A \rightarrow A$, $\tau: A \rightarrow A$. We can introduce group on $\mathrm{Hom}(X, A)$ for any object $X$.

Morphisms $\mu$ and $\tau$ have to be such that the respective functions:
$\mathrm{Hom}(X, A) \times \mathrm{Hom}(X, A) \simeq \mathrm{Hom}(X, A \times A) \rightarrow \mathrm{Hom}(X, A)$, $\mathrm{Hom}(X, A) \rightarrow \mathrm{Hom}(X, A)$ satisfy group axioms. The arrows are defined in the natural way.

I think, that 'in the natural way', means that for $f, g \in \mathrm{Hom}(X, A)$ we have $f \cdot g = \mu \circ f \times g$, where $\times$ denotes product of morphisms (which exists due to the definition of product in category). And $f^{-1} = \tau \circ f$.

One problem is, what properties has to satisfy $\mu$ and $\tau$ in order to induce group structure.

But i don't want to think on this right now.

2. The second way, is to suppose we have group structure on each $\mathrm{Hom}(X, A)$ for given $A$ and any $X$. Of course the groups have to be related somehow. And the relation is, that for any morphism $\mathrm{Hom}(Y, X)$ the respective arrow $\mathrm{Hom}(X, A) \rightarrow \mathrm{Hom}(Y, A)$ is a group homomorphism.

Finally the question is. We have group structure given in the way 2, and we want to define group structure in the way 1. It is, we want to define morphisms $\mu$ and $\tau$ in some way.
• November 21st 2011, 03:01 PM
albi
Re: Groups in Category Theory
My idea was to start with a group on $\mathrm{Hom}(A, A)$. Then we know, that the multiplication should be: $f \cdot g = \mu \circ f \times g$ for some $\mu$.

On the other hand we have to use the fact that the groups on $\mathrm{Hom}(X, A)$ are related... Hmm, I think I've just got one idea (looks like writing this thing on the forum is helpful..., but I will think on this tomorrow)
• November 21st 2011, 09:11 PM
Drexel28
Re: Groups in Category Theory
I'm sorry this is hard to read. what precisely are you trying to come up with? Are you trying to define, in general, a group operation on the hom-sets? Or, are you looking for some notion that, when satisfied, is useful, and has to do with the hom-sets being $\mathbb{Z}$-modules? If the former is true you can't always, in the sense that you can't make them into precisely the natural sort of group that distributes over composition. In other words, you can't always make them into a group in such a way that the category becomes preadditive.
• November 22nd 2011, 11:05 AM
albi
Re: Groups in Category Theory
No, no. I think it has nothing to do with preaddtive categories or abelian groups. But writing this whole thing down, helped me to understand a bit of it, and I know to solve it.

So we have a group structures (not necessarly abelian) on $\mathrm{Hom}(X, A)$ where $A$ is fixed. In particular we have a group structure for $\mathrm{Hom}(A, A)$. Let us denote its group operations by $\cdot$ and ${}^{-1}$.

I am looking for a morhpism $\mu$ such that the multiplication is $f \cdot g = \mu \circ f \times g$. I want to express it by the group operations, neutral element, etc.

What is important to notice, that $\pi_1 \times \pi_2 = \mathrm{id}_{A \times A}$ where $\pi_i: A \times A \rightarrow A$ are projections. So substituting them for $f$ and $g$ we get $\mu = \pi_1 \cdot \pi_2$.

Which is not true. But it was the thing I came up yesterday. The problem is that the projections are not good morphisms. So I still don't know.
• November 22nd 2011, 03:00 PM
Deveno
Re: Groups in Category Theory
i think what you are trying to do is define "point-wise" multiplication. for example, if our category is Top, then Hom(X,A) is continuous functions from X to A.

so we can make this into a group by taking:

$(f*g)(x) = f(x)g(x)$
$f^{-1}(x) = (f(x))^{-1}$
$e_{\mathrm{Hom}(X,A)}(x) = e_A$.
• November 22nd 2011, 03:42 PM
albi
Re: Groups in Category Theory
Yes. In Top we get topological groups, for smooth manifolds we get Lie groups (and the operations on hom-sets are poinwise). But does it help me in the general case?
• November 22nd 2011, 07:34 PM
Drexel28
Re: Groups in Category Theory
Quote:

Originally Posted by albi
Yes. In Top we get topological groups, for smooth manifolds we get Lie groups (and the operations on hom-sets are poinwise). But does it help me in the general case?

Ah, perhaps what you mean then is a group object in a category. This is slightly more general than you are considering, but should cover your case nicely, no? Sorry if I have double misunderstood.
• November 23rd 2011, 05:06 AM
Deveno
Re: Groups in Category Theory
i think you're talking about the same thing. if $X$ lies in a category $\mathcal{C}$, and we have a functor $F:\mathcal{C} \to \bold{Grp}$ given by: $F(X) = \mathrm{Hom}(X,A)$, then $A$ is a group object in $\mathcal{C}$.

clearly, since $F$ is a functor, we also have a group morphism $F(f)$ for every $f \in \mathrm{Hom}(X,Y)$, in $\mathrm{Hom}(\mathrm{Hom}(X,A),\mathrm{Hom}(Y,A))$.

what does this mean?

we have a morphism $\mu:\mathrm{Hom}(X,A) \times \mathrm{Hom}(X,A) \to \mathrm{Hom}(X,A)$, a morhpism $\eta:1 \to \mathrm{Hom}(X,A)$, and a morphism $\iota:\mathrm{Hom}(X,A) \to \mathrm{Hom}(X,A)$ such that:

$\mu \circ (\mu \times \mathrm{id}_{\mathrm{Hom}(X,A)}) = \mu \circ (\mathrm{id}_{\mathrm{Hom}(X,A)} \times \mu)$
$\mu \circ (\eta \times \mathrm{id}_{\mathrm{Hom}(X,A)}) = \pi_1,\ \mu \circ (\mathrm{id}_{\mathrm{Hom}(X,A)} \times \eta) = \pi_2$
$\mu \circ (\mathrm{id}_{\mathrm{Hom}(X,A)} \times \iota) \circ \Delta = \mu \circ (\iota \times \mathrm{id}_{\mathrm{Hom}(X,A)}) \circ \Delta = \eta \circ 1$

where:

$\pi_1,\ \pi_2$ are the canonical projections on the product, $\Delta$ is the diagonal map, and $1$ is the unique map: $1: \mathrm{Hom}(X,A) \to 1$

now, i think albi's question is this:

given only $F$, how do we construct $\mu,\ \eta,$ and $\iota$? and the answer is: "point-wise in A".
• November 26th 2011, 11:28 AM
albi
Re: Groups in Category Theory
I don't understand anything after "what does it mean?".

In particular, I don't understand how
$\mathrm{id}_A \times \mu$ makes sense. Because the property of of product is such that, when we have $f_i : X \rightarrow A$ there exists unique $f_1 \times f_2: X \rightarrow A \times A$ such that $f_i$ factorizes on f and $\pi_i$. But $\mathrm{id}_A$ and $\mu$ have diffrent "domains" (or whatever is the name in category theory).
• November 26th 2011, 12:57 PM
Deveno
Re: Groups in Category Theory
a categorical product does not have involve a single object. that is, you can have A x B and not just A x A. in which case, if you have f: X→Y and g:Z→W you can have f x g:X x Z → Y x W. for example, if we are talking about sets, (f x g)(x,y) = (f(x),g(z)). the "coordinates" of f x g are simply $\pi_1 \circ (f \times g) = f$ and $\pi_2 \circ (f \times g) = g$, and we can use this formulation in an arbitrary category that has products.

so, explicitly, $(\mathrm{id}_A \times \mu)(a,b,c) = (a,\mu(b,c))$ for those categories where morphisms are maps of some sort.
• November 26th 2011, 03:01 PM
albi
Re: Groups in Category Theory
I still don't understand how it should work (unless products are isomorphic to coproducts, or something :P). In your example cannot be $\pi_1 \circ (f \times g) = f$ because LHS is $X \times Z \rightarrow Y$ and RHS is $X \rightarrow Y$.
• November 26th 2011, 06:07 PM
Deveno
Re: Groups in Category Theory
well, that's true...the RHS should more properly be something like $f \circ \pi_1$ to match up the domains (i was thinking of just the values, d'oh).

but i think you're missing the point: given something in Hom(X,Y) and something in Hom(Z,W), there is a natural way to create something in Hom(X x Y, Z x W).
• November 28th 2011, 07:37 AM
albi
Re: Groups in Category Theory
I get it now. The trick is to compose our morhphisms with respective $\pi$'s. Then we get morhphisms which have the same "domain". So there exists unique (and hence natural) $\times$.

I will have some more questions on this, when I get home and have time to think about it...

PS But i have to admit, that it is the reason why I could not derive properties of $\mu$. But since I can $\times$ morphisms like that, things come easier for me.
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