Let $0 \rightarrow M' \rightarrow^f M \rightarrow^g M'' \rightarrow 0$ be a exact sequence. Then
If there is a homomorphism $\psi: M \rightarrow M' $ such that $\psi \circ f = I_{M'}$ then there is a homomorphism $\phi : M'' \rightarrow M$ such that $g \circ \phi = I_{M''}.$
Since we have a exact sequence then $f$ is injective, $g$ is surjective and $f(M') = g^{-1}(0)$.
I was able to show that $M = \psi^{-1}(0) \oplus f(M')$.
I am having trouble to define $\phi$. I know that for every $x \in M'' $, since g is surjective, there is a $y \in M$ such that $g(y)=x$. Now we have that $y = z + w$, where $z \in \psi^{-1}(0)$ and $w \in f(M')$. I got stuck here, I tried a few paths but no success.
Any help would be nice.
Hint: Try just defining $\phi(x)=z$. (You will have to check that this is well-defined; in other words, it does not depend on the choice of $y$ such that $g(y)=x$.)