Morphism extends iff image trought connecting map is zero I am studying the $\operatorname{Ext}$ functor, and I've found the following problem. It is from Assem's Àlgebres et Modules. Assume that all modules are over a ring $R$.

Let $0 \rightarrow M' \rightarrow M \rightarrow M' \rightarrow 0$ be a short exact sequence. Show that a morphism $f': M' \rightarrow N$ extends to a morphism $M \rightarrow N$ if, and only if, the image of $f'$ trought the connecting morphism $\operatorname{Hom}_R(M',N) \rightarrow \operatorname{Ext}_A^1(M'',N)$ is zero.

What I've done: For the $(\Rightarrow)$ implication, since we have the diagram
$\require{AMScd}$
\begin{CD}
0@> >> M' @>{u}>> M @>{v}>> M'' @> >> 0\\
@. @V{f'}VV \\
@. N
\end{CD}
in which the line is exact, we can complete it to an commutative diagram
$\require{AMScd}$
\begin{CD}
0@> >> M' @>{u}>> M @>{v}>> M'' @> >> 0\\
@. @V{f'}VV @V{g'}VV @V{1}VV \\
0 @> >> N @>{u'}>> N' @>{v'}>> M'' @> >> 0
\end{CD}
in which both lines are short exact sequences, and $N'$ is the pushout of $(u,f')$. From here, I think that the next step might involve some extensions classes, but I do not know how to proceed.
I was able to do little to nothing with this problem. One of the main issues is that I do not concretely understand the $\operatorname{Ext}$ functor.
If someone could give me some ideas on how to proceed, it would help a lot.
 A: Let $I$ be an injective module containing $N$, and let $Q$ denote the quotient $I/N$.  By the injective property of $I$ we may construct a map $f$ (hence also $f''$) such that the following diagram commutes.
$\require{AMScd}$
\begin{CD}
0@> >> M' @>{u}>> M @>{v}>> M'' @> >> 0\\
@. @V{f'}VV @VfVV @Vf''VV \\
0@> >> N @>{\iota}>> I @>{\epsilon}>> Q @> >> 0
\end{CD}
Now $f'\in \operatorname{Hom}_R(M',N)$, and its image under the connecting morphism is the class represented by $f''$ in $\operatorname{Ext}_R^1(M'',N)$.
This class is trivial if and only if $f''$ factors through $\epsilon$:$$f''=\epsilon h,$$
for some $h\colon M''\to I$.
Given such a map $h$, we have $\epsilon(f-hv)=0$, so $$f-hv=\iota g $$ for some map $g\colon  M\to N$.
It follows that $\iota(f'-gu)=0$, so $f'=gu$ as required.
Conversely, if we have a map $g\colon  M\to N$ such that $f'=gu$, then $(f-\iota g)u=0$, so $$f-\iota g= hv,$$ for some map $h\colon M''\to I$.
It follows that $(f''-\epsilon h)v=0$, so $f''=\epsilon h$ and $f''$ represents the trivial class in $\operatorname{Ext}_R^1(M'',N)$, as required.

Here is a brief explanation for why the element of $\operatorname{Ext}_R^1(M'',N)$ represented by the cocycle $f''$ is the same as the one you constructed in your attempt.
Consider the following commutative diagram, where the bottom right square is a pull-back diagram:
$\require{AMScd}$
\begin{CD}
0@> >> M' @>{u}>> M @>{v}>> M'' @> >> 0\\
@. @V{f'}VV @Vf\times vVV @V1_{M''}VV \\
0@> >> N @>{\iota\times 0}>> I\times_{Q}M'' @>{p_{M''}}>> M'' @> >> 0\\
@. @V{1_N}VV @Vp_IVV @Vf''VV \\
0@> >> N @>{\iota}>> I @>{\epsilon}>> Q @> >> 0
\end{CD}
In general, given a cocycle we can recover an extension by taking pull-backs as above, so the middle row is the extension represented by $f''$.
It is an easy check that the top left square is a pushout diagram, so the middle row is just your short exact sequence:
\begin{CD}
0 @> >> N @>{u'}>> N' @>{v'}>> M'' @> >> 0
\end{CD}
A: It may be more helpful to give you an answer along the lines begun in your attempt:
You are correct that the image of $f'$ in $\operatorname{Ext}_R^1(M'',N)$ is represented by your extension:
\begin{CD}
0 @> >> N @>{u'}>> N' @>{v'}>> M'' @> >> 0
\end{CD}
I have added a brief explanation to the bottom of my previous answer, for why this is the same as what I stated in my previous answer.
Now an extension represents the zero element in $\operatorname{Ext}_R^1(M'',N)$ precisely when it splits:  In this case when there is a map $s\colon N'\to N$ such that $su'=1_N$.
Suppose such a map $s$ exists.  Then $sg'\colon M \to N$ is the required extension of $f'$.  Here $g'$ is the map you defined in your commutative diagram.
Conversely given an extension of $f'$ to a morphism $f\colon M \to N$, we may define the splitting $s\colon N'\to N$ by $$s(n,m)=n+f(m).$$
