# The Unapologetic Mathematician

## 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 $\mathcal{C}$ is a category with a zero object, kernels, and cokernels, then any arrow $f$ of $\mathcal{C}$ has a factorization $f=m\circ q$, where $m=\mathrm{Ker}(\mathrm{Cok}(f))$. This holds because $\mathrm{Cok}(f)\circ f=0$, and by the universal property of kernels there is a unique $q$ with $f=m\circ q$.

On the other hand, if we have another factorization $f=m'\circ q'$ with $m'$ also a kernel, then we have a unique $t$ so that the following diagram commutes.

That is, the canonical factorization described above is in some sense the universal such factorization. Furthermore, if $\mathcal{C}$ has equalizers and every monic in $\mathcal{C}$ is a kernel, then $q$ 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 $q$ and a monic $m$.

We prove this by considering the following diagram:

We draw $p'=\mathrm{Cok}(m')$. Since $m'$ is itself a kernel, we see that $m'=\mathrm{Ker}(p')$. We also draw $p=\mathrm{Cok}(m)=\mathrm{Cok}(f)$. Now $p'\circ m'=0$, so $p'\circ f=p'\circ m'\circ q'=0$. By the universal property of $p$ , there is a unique $w$ so that $p'=w\circ p$. And then $p'\circ m=w\circ p\circ m=0$, so $m$ factors uniquely through $m'$ — there is a unique $t$ with $m=m'\circ t$. Furthermore, $m'\circ q'=m\circ q=m'\circ t\circ q$, and so since $m'$ is monic we have $q'=t\circ q$. This proves that the first diagram given above commutes.

Now we have to show that $q$ is epic under the additional hypotheses. Let’s say that we have a parallel pair of arrows $r$ and $s$ with $r\circ q=s\circ q$. then $q$ factors uniquely through the equalizer $\mathrm{Equ}(r,s)$$q=e\circ q'$ for some unique $q'$. Then $f=m\circ q=m\circ e\circ q'$. Now $m'=me$ is monic, so (by the new hypotheses) it’s a kernel. By the first part of the theorem we have a unique $t$ with $m=m'\circ t=m\circ e\circ t$, and thus $1=e\circ t$. Since the monic $e$ has a right inverse, it’s an isomorphism. Since $e$ was picked as the equalizer of $r$ and $s$, $r=s$. And so $q$ is epic.

September 25, 2007 - Posted by | Category theory

## 8 Comments »

1. […] Coimages, and Exactness By the first isomorphism theorem, we know that any morphism in an abelian category factorizes as with , and is epic. Since is […]

Pingback by Images, Coimages, and Exactness « The Unapologetic Mathematician | September 26, 2007 | Reply

2. […] We already said that (also ) is a abelian category. Now in any abelian category we have the first isomorphism theorem. […]

Pingback by The Rank-Nullity Theorem « The Unapologetic Mathematician | June 27, 2008 | Reply

3. 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 | Reply

4. 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 $\mathbf{Set}$, 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 | Reply

5. 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, $\mathbf{Set}$ 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 | Reply

6. […] it’s a very concrete representation of the first isomorphism theorem. Every transformation is decomposed into a projection, an isomorphism, and an inclusion. But here […]

Pingback by The Meaning of the SVD « The Unapologetic Mathematician | August 18, 2009 | Reply

7. […] for the singular value decomposition, what we’ll end up with is essentially captured in the first isomorphism theorem, but we’ll be able to be a lot more explicit about how to find the right bases to simplify […]

Pingback by Decompositions Past and Future « The Unapologetic Mathematician | August 24, 2009 | Reply

8. […] may not be Abelian, but it has a zero object, kernels, and cokernels, which is enough to get the first isomorphism theorem, just like for rings. Specifically, if is any homomorphism of Lie algebras then we can factor it […]

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