equivalent conditions for an exact sequence to split Let $0 \rightarrow M' \xrightarrow{f} M \xrightarrow{g} M'' \rightarrow 0$ by an exact sequence of modules. the following conditions are equivalent

*

*There exists a homomorphism $\phi: M'' \rightarrow M$ such that $g \circ \phi = $ id.


*There exists a homomorphism $\psi: M \rightarrow M'$ such that $\psi \circ f = $ id.
$(1) \implies (2)$. I try to show $f$ is injective and surjetive. Since the sequence $0 \rightarrow M' \xrightarrow{f} M$ is exact, $f$ is injective. Since $g$ has an inverse, $\ker(g) = 0$. Thus, since the sequence $M' \xrightarrow{f} M \xrightarrow{g} M''$ is exact, $f$ maps identically to $0$???
This last step I'm having some trouble getting around. If $f$ maps to $0$ it can't be surjective unless $M$ only contains $0$.
 A: $\require{AMScd}$
We will show that your statements 1. and 2. are equivalent to the following:


*There exists an isomorphism $\mu : M \to M' \oplus M''$ such that the following diagram with exact rows commutes
\begin{CD}
0 @>>> M' @>f>> M @>g>> M'' @>>> 0\\
 @. @V\text{Id}_{M'}VV  @V\mu VV  @VV\text{Id}_{M''}V\\
0 @>>> M' @>i>> M' \oplus M'' @>p>> M'' @>>> 0
\end{CD}
Here, $i$ and $p$ are the obvious injection and projection respectively.
(3. $\implies$ 2.) Define $\psi := \text{pr}_{M'} \circ \mu$. Here, $\text{pr}_{M'} : M' \oplus M'' \to M'$ is defined by $(m',m'') \mapsto m'$. Then for $m' \in M'$ we have
$$
(\psi \circ f)(m') = \text{pr}_{M'}(\mu(f(m'))) = \text{pr}_{M'}(i(m')) = \text{pr}_{M'}(m',0) = m',
$$
where we used that the left square in the diagram commutes.
(3. $\implies$ 1.) Define $\phi(m'') := \mu^{-1}(0,m'')$ for $m'' \in M''$. Then we have
$$
(g \circ \phi)(m'') = g(\mu^{-1}(0,m'')) = p(0,m'')=m'',
$$
where again we used that the right square commutes, and that $\mu$ is bijective.
(2. $\implies$ 3.) Let $\mu(m) := (\psi(m),g(m))$. You can check that the diagram does indeed commute. By the five lemma, $\mu$ is an isomorphism, since $\text{Id}$ is always bijective.
(1. $\implies$ 3.) Define $\nu : M' \oplus M'' \to M$ by $\nu(m',m'') = f(m')+\phi(m'')$. You can check that the following diagram then commutes:
\begin{CD}
0 @>>> M' @>f>> M @>g>> M'' @>>> 0\\
 @. @A\text{Id}_{M'}AA  @A\nu AA  @AA\text{Id}_{M''}A\\
0 @>>> M' @>i>> M' \oplus M'' @>p>> M'' @>>> 0
\end{CD}
Again, by the five lemma we have that $\nu$ is an isomorphism, so 3. follows.
Now we have shown that 1. and 2. are equivalent to 3., so in particular 1. and 2. are equivalent. Hope this helps!
