# Problem of the Week #62 - August 5th, 2013

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#### Chris L T521

##### Well-known member
Staff member
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:
$0\rightarrow U \xrightarrow{\psi} V \xrightarrow{\phi} W \rightarrow 0.$
Define
$S = \{\sigma \in \text{Hom}_K(W,V) : \phi \circ \sigma = \text{Id}_W\}.$
(An element of S is called a splitting of the short exact sequence). Prove that there exists a bijection from $\text{Hom}_K(W,U)$ to $S$.

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#### Chris L T521

##### Well-known member
Staff member
No one answered this week's question. You can find my solution below.

Since the sequence$0\rightarrow U \xrightarrow{\psi} V \xrightarrow{\phi} W \rightarrow 0$
is exact, then we have
$V = \ker\ \phi \oplus F,$
where $F \cong W$. We denote the projections as $p_1 : V \rightarrow \ker\ \phi$ and $p_2 : V \rightarrow F$. While $\psi$ is injective, then $\exists \varphi \in \text{Hom}_K (V, U)$ such that $\varphi\circ\psi = Id_U.$

Let $\alpha : W \rightarrow U$ be
$\alpha = \varphi|_{\ker\ \phi} \circ p_1 \circ \sigma.$
It is easy to see that $\varphi$, $p_1$ and $\sigma$ is $K$-linear, thus $\alpha \in \text{Hom}_K(W,U)$ and $\alpha$ is decided by $\sigma$ completely.

Let
$h: S \rightarrow \text{Hom}_K(W,U);\ \ h(\sigma) = \varphi|_{\ker\ \phi} \circ p_1 \circ \sigma,$
then $h$ is well-defined.

If $\exists \sigma_1, \sigma_2$ such that $h(\sigma_1) = h(\sigma_2)$, i.e.
$\varphi|_{\ker\ \phi} \circ p_1 \circ \sigma_1 = \varphi|_{\ker\ \phi} \circ p_1 \circ \sigma_2.$

Since $\varphi|_{\ker\ \phi} : \ker\ \phi \rightarrow U$ is an isomorphism, then
$(\psi \circ \varphi|_{\ker\ \phi}) \circ p_1 \circ \sigma_1 = (\psi \circ \varphi|_{\ker\ \phi}) \circ p_1 \circ \sigma_2,$
i.e. $p_1 \circ \sigma_1 = p_1 \circ \sigma_2.$

Since $\sigma = (p_1\circ \sigma, p_2\circ \sigma)$, then
$\sigma_1 - \sigma_2 = (p_1\circ (\sigma_1-\sigma_2), p_2\circ (\sigma_1-\sigma_2)).$
Therefore,
$\phi\circ (\sigma_1 - \sigma_2) = \phi\circ p_2\circ (\sigma_1-\sigma_2).$
i.e.
$0 = \text{Id}_W - \text{Id}_W = \phi\circ p_2\circ (\sigma_1-\sigma_2).$
Since $\phi\circ p_2 \circ (\sigma_1-\sigma_2) = \phi|_{F}\circ p_2 \circ (\sigma_1-\sigma_2)$ and $\phi|_{F}$ is injective, therefore $p_2 \circ (\sigma_1-\sigma_2) = 0$. Hence
$(p_1\circ \sigma_1, p_2\circ \sigma_1) = (p_1\circ \sigma_2, p_2\circ \sigma_2).$
i.e. $\sigma_1 = \sigma_2$. Now we have proved $h$ is injective.

Now $\forall \alpha \in \text{Hom}_k(W,U)$, let $\sigma = \psi \circ \alpha$. Then
$h(\sigma) = \varphi|_{\ker\ \phi} \circ p_1 \circ \sigma = \varphi|_{\ker\ \phi} \circ p_1 \circ \psi \circ \alpha.$

Since $\psi \circ \alpha : \text{Hom}_K(W,\ker\ \phi)$, then
$h(\sigma) = \varphi \circ \psi \circ \alpha = \text{Id}_U \circ \alpha = \alpha.$
Therefore, $h$ is a surjective.

Combining both of the parts, we conclude that $h: S \rightarrow \text{Hom}_K (W,U)$ is bijective.

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