The First Isomorphism Theorem (for Abelian Categories)
We had versions of the first isomorphism theorem for groups, rings, and modules. Now we’ll do it in a more general setting. We’re going to use it in our study of abelian categories, but it works in all three of the above cases.
As we said last time we talked about abelian categories, we really just use that our categories all have a zero object, kernels, and cokernels. We’ll work with that for now, and note that these properties hold in particular for abelian categories.
Now, if is a category with a zero object, kernels, and cokernels, then any arrow of has a factorization , where . This holds because , and by the universal property of kernels there is a unique with .
On the other hand, if we have another factorization with also a kernel, then we have a unique so that the following diagram commutes.
That is, the canonical factorization described above is in some sense the universal such factorization. Furthermore, if has equalizers and every monic in is a kernel, then is epic. In particular, these hypotheses are all satisfied for abelian categories. This is the isomorphism theorem — that every arrow factorizes essentially uniquely as the composition of an epic and a monic .
We prove this by considering the following diagram:
We draw . Since is itself a kernel, we see that . We also draw . Now , so . By the universal property of , there is a unique so that . And then , so factors uniquely through — there is a unique with . Furthermore, , and so since is monic we have . This proves that the first diagram given above commutes.
Now we have to show that is epic under the additional hypotheses. Let’s say that we have a parallel pair of arrows and with . then factors uniquely through the equalizer — for some unique . Then . Now is monic, so (by the new hypotheses) it’s a kernel. By the first part of the theorem we have a unique with , and thus . Since the monic has a right inverse, it’s an isomorphism. Since was picked as the equalizer of and , . And so is epic.
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I was wondering if there’s a way to do this in more general categories – for example, I was told about a “First Isomorphism Theorem for Sets” that goes as follows:
In $\mathbf{Set}$, consider a map $f \colon X \to Y$, and consider $Z$ given by $X / \sim$ where $\sim$ is the equivalence relation given by $a \sim b$ if $f(a) = f(b)$.
Then the First Isomorphism Theorem tells us that $\tilde{f} \colon Z \to \mathrm{im}(f)$ is an isomorphism. Is there anyway to put this in a more categorical setting?
Comment by Sam | February 17, 2009 |
I’m sorry, it seems I misunderstood how LaTeX works on WordPress. The above post should have read:
I was wondering if there’s a way to do this in more general categories – for example, I was told about a “First Isomorphism Theorem for Sets” that goes as follows:
In , consider a map , and consider given by where is the equivalence relation given by if .
Then the First Isomorphism Theorem tells us that is an isomorphism. Is there anyway to put this in a more categorical setting?
Comment by Sam | February 17, 2009 |
Yes, if you consider carefully what properties of an Abelian category we’re using here. As I note above, this proof just uses the fact that we have a zero object, kernels, and cokernels.
Now, doesn’t have a zero object, so there are no kernels and cokernels. But there are equalizers and coequalizers, so that gives you something to start gnawing at.
Comment by John Armstrong | February 17, 2009 |
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