The Unapologetic Mathematician

The Splitting Lemma

Evidently I never did this one when I was talking about abelian categories. Looks like I have to go back and patch this now.

We start with a short exact sequence:

$\mathbf{0}\rightarrow A\xrightarrow{f}B\xrightarrow{g}C\rightarrow\mathbf{0}$

A large class of examples of such sequences are provided by the split-exact sequences:

$\mathbf{0}\rightarrow A\rightarrow A\oplus C\rightarrow C\rightarrow\mathbf{0}$

where these arrows are those from the definition of the biproduct. But in this case we’ve also got other arrows: $A\oplus C\rightarrow A$ and $C\rightarrow A\oplus C$ that satisfy certain relations.

The lemma says that we can go the other direction too. If we have one arrow $h:B\rightarrow A$ so that $h\circ f=1_A$ then everything else falls into place, and $B\cong A\oplus C$. Similarly, a single arrow $h:C\rightarrow B$ so that $g\circ h=1_C$ will “split” the sequence. We’ll just prove the first one, since the second goes more or less the same way.

Just like with diagram chases, we’re going to talk about “elements” of objects as if the objects are abelian groups. Of course, we don’t really mean “elements”, but the exact same semantic switch works here.

So let’s consider an element $b\in B$ and write it as $(b-f(h(b)))+f(h(b))$. Clearly $f(h(b))$ lands in $\mathrm{Im}(f)$. We can also check

$h(b-f(h(b)))=h(b)-h(f(h(b)))=h(b)-h(b)=0$

so $b-f(h(b))\in\mathrm{Ker}(h)$. That is, any element of $B$ can be written as the sum of an element of $\mathrm{Im}(f)$ and an element of $\mathrm{Ker}(h)$. But these two intersect trivially. That is, if $b=f(a)$ and $h(b)=0$ then $0=h(f(a))=a$, and so $b=0$. This shows that $B\cong\mathrm{Ker}(h)\oplus\mathrm{Im}(f)$. Thus we can write every $b$ uniquely as $b=f(a)+k$.

Now consider an element $c\in C$. By exactness, there must be some $b\in B$ so that $c=g(b)=g(f(a)+k)=g(f(a))+g(k)$. That is, we have a unique $k\in\mathrm{Ker}(h)$ with $g(k)=c$. This shows that $C\cong\mathrm{Ker}(h)$. It’s straightforward to show that also $A\cong\mathrm{Im}(f)$. Thus we have split the sequence: $B\cong A\oplus C$.

June 25, 2008 Posted by | Category theory | 6 Comments