# The Unapologetic Mathematician

## Exact sequences split

Now that we know the splitting lemma, we can show that every short exact sequence of vector spaces splits!

To see this, we’ll need to refine an earlier result. Remember how we showed that every vector space has a basis. We looked for maximal linearly independent sets and used Zorn’s lemma to assert that they existed.

Here’s how we’re going to refine this result: start with a collection $S$ of linearly independent vectors. Then we don’t just look for any maximal collection, but specifically for a maximal collection containing $S$. Clearly if we have a linearly independent set $B$ containing $S$ which is maximal among such sets, it is also maximal among all linearly independent sets — it is a basis. On the other hand, the previous argument (with Zorn’s lemma) says that such a maximal linearly independent set must exist.

What does this mean? It says that any linearly independent set can be completed to a basis. If we start with the empty set (which is trivially linearly independent) then we get a basis, just like before. So we recover the same old result as before.

But look what we can do now! Take a short exact sequence

$\mathbf{0}\rightarrow U\xrightarrow{S}V\xrightarrow{T}W\rightarrow\mathbf{0}$

and pick any basis $\left\{e_i\right\}_{i\in\mathcal{I}}$ of $U$ (notice that we’re using a generic, possibly infinite, index set). Now hit this basis with $S$ to get $\left\{f_i\right\}_{i\in\mathcal{I}}$ with $f_i=S(e_i)$. I say that this is a linearly independent set in $V$.

Why is this? Well, let’s say that there’s a linear combination $r^if_i=0$ (only finitely many of the $r^i$ can be nonzero here). This linear combination is in the image of $S$, since we can write it as $r^iS(e_i)=S(r^ie_i)$. But exactness tells us that $S$ is injective, and so we have $r^ie_i=0$. But then all the $r^i$ have to vanish, since the $e_i$ form a basis!

So we’ve got a linearly independent set $\left\{f_i\right\}_{i\in\mathcal{I}}$ in $V$. We now complete this to a maximal linearly independent set $\left\{f_i\right\}_{i\in\mathcal{I}'}$ with $\mathcal{I}\subseteq\mathcal{I}'$. This is now a basis for $V$, which contains the image of all the basis elements from $U$.

Now turn this around and define a linear transformation $S':V\rightarrow U$ by specifying its values on this basis. Set $S'(f_i)=e_i$ for $i\in\mathcal{I}$ and $S'(f_i)=0$ for $i\in\mathcal{I}'\setminus\mathcal{I}$. Then the composition $S'\circ S$ is the identity on $U$ (check it on our basis), and so the sequence splits, as we said.

Notice here that all the Zorniness only matters for infinite-dimensional vector spaces. Everything we’ve done here works in $\mathbf{FinVec}(\mathbb{F})$ without ever having to worry about such set-theoretic problems. However, given my politics I have no problem with using the Axiom of Choice when push comes to shove. I’m just pointing this out in case you’re the squeamish sort.

June 26, 2008 Posted by | Algebra, Linear Algebra | 11 Comments