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 John Armstrong | Category theory

6 Comments »

So …. I’m not sure I quite fathomed the semantic switch in grabbing elements. What you -REALLY- mean by something like b – f(h(b)) is looking at the -morphism- Id-f.h is it?

And then equality, and being 0, and all the arithmetic just works out?

Comment by Mikael Vejdemo Johansson | June 25, 2008 | Reply
Go read the link. The switch is that an “element” $b\in B$ is really an arrow $b:X\rightarrow B$ from an arbitrary object $X$ . Then you have to switch other definitions, like what’s meant by the $=$ sign to compensate.

Really, the only thing new here is that I’m using the abelian group structure on the hom-sets where I didn’t need it for the other lemmas I proved with the “semantic” diagram chases.

Comment by John Armstrong | June 25, 2008 | Reply
[…] sequences split Now that we know the splitting lemma, we can show that every short exact sequence of vector spaces […]

Pingback by Exact sequences split « The Unapologetic Mathematician | June 26, 2008 | Reply
[…] The exact sequence splits. Indeed, we can lift a basis of to elements of by surjectivity; then define a map from that […]

Pingback by Generic freeness II « Delta Epsilons | July 31, 2009 | Reply
Hi!

What books reference the Splitting Lemma?

Thanks!

Comment by JoseBrox | April 13, 2011 | Reply
Pretty much any homological algebra book should talk about it.

Comment by John Armstrong | April 13, 2011 | Reply

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