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

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Chris L T521

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Jan 26, 2012
995
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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
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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