A question about split short exact sequence of modules Let $0\longrightarrow A\stackrel{f}{\longrightarrow} C\stackrel{g}{\longrightarrow} B\longrightarrow 0$ be a split short exact sequence of modules.
That is, 
there exists $\alpha:C\to A$ such that $\alpha\circ f=1_{A}$
or there exists $\beta:B\to C$ such that $g\circ \beta=1_{B}$.
We have 
\begin{eqnarray*}
C &\cong& \text{Im }f\oplus \ker{\alpha}\\
&\cong& \ker{g}\oplus \text{Im }\beta.
\end{eqnarray*}
It seems $\ker{\alpha}=\text{Im }\beta$ because $\text{Im }f=\ker{g}$.
But I can't prove it and can't find a counterexample.
Is it ($\ker{\alpha}=\text{Im }\beta$) true?
(There is an example in group theory
which shows that $A\oplus B\cong A\oplus C$ doesn't imply $B=C$.
For example, $\Bbb{Z}_2\oplus \Bbb{Z}_2\cong \langle (1,0)\rangle\oplus \langle (0,1)\rangle\cong \langle (1,0)\rangle\oplus \langle (1,1)\rangle$.)
 A: Note that if the sequence splits, then there are typically many sections of $g$. If $\beta \colon B \to C$ satisfies $g\circ \beta = 1_B$, and $\delta \colon B \to \ker g$ is any homomorphism, then $\beta + \delta$ is also a section of $g$:
$$g\bigl((\beta + \delta)(b)\bigr) = g\bigl(\beta(b) + \delta(b)\bigr) = g\bigl(\beta(b)\bigr) + g\bigl(\underbrace{\delta(b)}_{\in \ker g}\bigr) = g\bigl(\beta(b)\bigr) = b.$$
Conversely, if $\beta_1,\beta_2$ are sections of $g$, then $\operatorname{Im} (\beta_1 - \beta_2) \subset \ker g$, so the above construction yields all sections of $g$.
And any two different sections of $g$ have different image. If $\beta_1,\beta_2$ are different sections of $g$, and $b\in B$ is such that $\beta_1(b) \neq \beta_2(b)$, then $\beta_1(b) \in \operatorname{Im} \beta_1 \setminus \operatorname{Im} \beta_2$. For if there were $b'\in B$ with $\beta_1(b) = \beta_2(b')$, then $b = g\bigl(\beta_1(b)\bigr) = g\bigl(\beta_2(b')\bigr) = b'$, contradicting $\beta_1(b) \neq \beta_2(b)$.
Thus in general, we cannot expect $\ker \alpha = \operatorname{Im} \beta$.
For a concrete example, let $R$ be a (nontrivial) ring, and consider
$$0 \to R \xrightarrow{r \mapsto (r,0)} R \times R \xrightarrow{(r,s) \mapsto s} R \to 0$$
with $\alpha(r,s) = r$, so $\ker \alpha = \{0\} \times R$ and $\beta(s) = (s,s)$. The image of $\beta$ is the diagonal of $R\times R$, and $\ker \alpha \cap \operatorname{Im} \beta = \{(0,0)\}$.
However, if the sequence splits, then we can always choose $\alpha,\beta$ in such a way that $\ker \alpha = \operatorname{Im} \beta$.
If $\alpha$ is given, then
$$\bigl(g\lvert_{\ker \alpha}\bigr) \colon \ker \alpha \to B$$
is a module isomorphism, and we can choose $\beta = \bigl(g\lvert_{\ker \alpha}\bigr)^{-1}$ [and of course this is the only section of $g$ with image $\ker \alpha$].
If $\beta$ is given, then $p = \beta \circ g \colon C \to C$ is a projection, $p \circ p = \beta \circ (g\circ \beta)\circ g = \beta \circ 1_B \circ g = \beta \circ g = p$, and $\ker p = \ker g = \operatorname{Im} f$, so $\alpha = f^{-1}\circ (1_C - p)$ is a homomorphism that satisfies $\alpha \circ f = 1_A$ and $\ker \alpha = \ker (1_C - p) = \operatorname{Im} p = \operatorname{Im} \beta$. Of course $\alpha$ is uniquely determined by the conditions $\alpha \circ f = 1_A$ and $\ker \alpha = \operatorname{Im} \beta$, since $C = \operatorname{Im} f \oplus \operatorname{Im} \beta$.
