MHB Problem of the Week #113 - July 28th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Let $U$, $V$, and $W$ be three left $K$-vector spaces, and $\psi$, $\phi$ linear maps, fitting into a short exact sequence: $$0 \longrightarrow U \xrightarrow{~\psi~} V \xrightarrow{~\phi~} W \longrightarrow 0.$$
Define
$$S = \{\sigma \in \mathrm{Hom}_K(W,V) : \phi \circ \sigma = \text{Id}_W\}.$$
(An element of $S$ is called a splitting of the short exact sequence). Prove that there exists a bijection from $\mathrm{Hom}_K(W,U)$ to $S$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Euge and Opalg. You can find Opalg's solution below.

[sp]The key fact here is that if $Y$ is a $K$-vector subspace of $X$ then there exists a $K$-vector subspace $Z$ of $X$ such that $X$ is canonically isomorphic to the direct sum $Y\oplus Z$ (in category-theoretic terms, the category $\mathbf{Vect}_K$ admits coproducts). This is proved by taking a basis for $Y$, extending it to a basis for $X$ and defining $Z$ to be the subspace generated by all the additional vectors introduced into the basis. In general, when the spaces are not finite-dimensional, this will require the use of the axiom of choice.

Take $Y$ to be the subspace $\psi(U) \subseteq V$ and let $Z$ be a complementary subspace as above. Then $Y$ is isomorphic to $U$ (because $\psi$ is injective). Also, $Z$ is isomorphic to $W$ (because the restriction $\phi|_Z: Z\to W$ is surjective and its kernel is in $Y\cap Z = \{0\}$). Denote by $\rho: \phi(z) \mapsto z$ the inverse homomorphism from $W$ to $Z$.

Now suppose that $\theta \in \mathbf{Hom}_K(W,U)$. Then $\psi \circ \theta \in \mathbf{Hom}_K(W,V)$ and $\phi \circ (\psi \circ \theta) =(\phi \circ \psi) \circ \theta) = 0$ (by exactness at $Y$). Define $\sigma_\theta:W\to V$ by $\sigma_\theta(w) = (\psi\circ\theta)(w) + \rho(w)$. Then $(\phi\circ\sigma_\theta)(w) = (\phi \circ \psi \circ \theta)(w) + (\phi\circ \rho)(w) = 0+w=w.$ Therefore $\sigma_\theta \in S$. The map $\theta \mapsto \sigma_\theta$ is clearly injective because $\psi$ is injective.

Conversely, suppose that $\sigma \in S$. For $w\in W$, $\sigma(w) - \rho(w) \in \ker(\phi) = Y$. Define $\theta(w) = \psi^{-1}(\sigma(w) - \rho(w))$. Then $\theta \in \mathbf{Hom}_K(W,X)$ and $\sigma_\theta(w) = (\psi \circ \psi^{-1})(\sigma(w) - \rho(w)) + \rho(w) = \sigma(w) - \rho(w) + \rho(w) = \sigma(w)$. So $\sigma = \sigma_\theta$. Thus the map $\theta \mapsto \sigma_\theta$ is surjective, and consequently bijective, as required.[/sp]