# The Unapologetic Mathematician

## Short Exact Sequences

Last time we defined a short exact sequence in an abelian category $\mathcal{C}$ to be an exact sequence of the form $\mathbf{0}\rightarrow A\rightarrow B\rightarrow C\rightarrow\mathbf{0}$. These are the objects of a category $\mathbf{Ses}(\mathcal{C})$, whose morphisms are triples of arrows $f:A\rightarrow A'$, $g:B\rightarrow B'$, $h:C\rightarrow C'$ making the two squares commute in the diagram: The category is clearly enriched over $\mathbf{Ab}$.

As a first step into this category, we show that if $f$ and $h$ are both monic, then $g$ is monic as well. First let’s see how to do this in $\mathbf{Ab}$. We want to show that if $g(b)=0$ for $b\in B$ that $b=0$. Now $h(e(b))=e'(g(b))=0$, but we assumed $h$ to be monic, so $e(b)=0$ $b\in\mathbf{Ker}(e)$. Since the top row is exact this means $b\in\mathrm{Im}(m)$, and so there’s an $a\in A$ so that $m(a)=b$. Now $m'(f(a))=g(m(a))=g(b)=0$, but both $m'$ and $f$ are monic. Thus $a=0$, and $m(a)=b=0$ as well.

That’s the normal way to proceed, and we call it a “diagram chase”. Just start with an element on the diagram at $B$ and “chase” it around the diagram. But this only works if $\mathcal{C}$ is made up of structured sets, and we haven’t assumed that at all! We need to work a tiny bit more abstractly and look only at the arrows.

So let’s take $k:\mathrm{Ker}(g)\rightarrow B$. Now $h\circ e\circ k=e'\circ g\circ k=0$. Since $h$ is monic, we have $e\circ k=0$. And thus $k$ factors uniquely through $m=\mathrm{Ker}(e)$ as $k=m\circ k'$. Now $0=g\circ k=g\circ m\circ k'=m'\circ f\circ k'$. Since $m'$ and $f$ are assumed monic, $k=0$, and so $\mathrm{Ker}(g)=\mathbf{0}$ and $g$ is monic.

Either way, we can dualize the whole theorem to say that if $f$ and $h$ are epic, then $g$ is epic too. Together, these results are called the “short five lemma”.

Here’s another lemma, based on this diagram: The square on the right is a pullback, and I say that if $f$ is epic then $f'$ is too. On the left is the kernel $k$ of $f$, and I say that it factors through the kernel $k'$ of $f'$ to make the diagram commute.

Let’s just skip the chase and go directly to the general picture. As usual, pullbacks can be defined by products and equalizers. We get the product as $B\oplus D$, which then has two arrows to $C$: $f\circ\pi_1$ and $g\circ\pi_2$. And we find their equalizer by taking the kernel of their difference: $m:S\rightarrow B\oplus D$. Then $g'=\pi_1\circ m$ and $f'=\pi_2\circ m$.

Now under the assumption on $f$, we know that $f\circ\pi_1-g\circ\pi_2$ is epic. Indeed, if $h\circ(f\circ\pi_1-g\circ\pi_2)=0$ then $0=h\circ(f\circ\pi_1-g\circ\pi_2)\circ\iota_1=h\circ f$. But since $f$ is assumed epic, $h=0$. Supposing now that $u\circ f'=0$, we see that $u\circ\pi_2\circ m=0$, so $u\pi_2$ factors through the cokernel of $m$, which is $f\circ\pi_1-g\circ\pi_2$. That is, $u\circ\pi_2=u'\circ(f\circ\pi_1-g\circ\pi_2)$, and so $u'\circ f=u'\circ(f\circ\pi_1-g\circ\pi_2)\circ\iota_1=u\circ\pi_2\circ\iota_1=0$. Since $f$ is epic, $u'=0$, and so $u=u\circ\pi_2\circ\iota_2=u'\circ(f\circ\pi_1-g\circ\pi_2)=0$. Thus $f'$ is epic.

As for the other assertion, the pair of arrows $k:A\rightarrow B$ and $0:A\rightarrow D$ satisfy $f\circ k=g\circ0=0$, and so there is a unique arrow $k':A\rightarrow S$ by the universal property of the pullback. In particular, $f'\circ k'=0:A\rightarrow D$. On the other hand, given any other $v$ with $f'\circ v=0$ we have $f\circ g'\circ v=g\circ f'\circ v=0$, so $g'\circ v$ factors uniquely through $k$ as $g'\circ v=k\circ v'$. Then $g'\circ v=g'\circ(k'\circ v')$ and $f'\circ v=0=f'\circ(k'\circ v')$. But since the arrow factoring through a pullback is unique we must have $v=k'\circ v'$. So $k'$ really is the kernel of $f'$, as asserted.

September 27, 2007 Posted by | Category theory | 5 Comments