TITLE: How to prove $\mathrm{Im}(\mathrm{Ext}_R^1(g,A'))=\mathrm{Ker}(\mathrm{Ext}_R^1(f,A'))$
QUESTION [5 upvotes]: I'm reading MacLane's "Homology" and got stuck at the proof of the following fact.
Theorem. Let $E:0\xrightarrow{}A\xrightarrow{f}B\xrightarrow{g}C\xrightarrow{}0$ be a short exact sequence of left $R$-modules. Let $A'$ be a left $R$-module, then the sequence 
$$
\mathrm{Ext}_R^1(C,A')\xrightarrow{\mathrm{Ext}_R^1(g,A')}\mathrm{Ext}_R^1(B,A')\xrightarrow{\mathrm{Ext}_R^1(f,A')}\mathrm{Ext}_R^1(A,A')
$$
is exact. 
Attempt. Since $gf=0$, then
$$
\mathrm{Ext}_R^1(f,A')\mathrm{Ext}_R^1(g,A')=\mathrm{Ext}_R^1(gf,A')=0
$$
so $\mathrm{Im}(\mathrm{Ext}_R^1(g,A'))\subset\mathrm{Ker}(\mathrm{Ext}_R^1(f,A'))$. 
Now take coset $[E_1]\in\mathrm{Ker}(\mathrm{Ext}_R^1(f,A'))$, then $[E_1f]=\mathrm{Ext}_R^1(f,A')([E_1])=0$. This means that $E_1f$ splits, which is equivalent that $g_f$ is a retraction, $f_f$ - coretraction. In order to show that $[E_1]\in\mathrm{Im}(\mathrm{Ext}_R^1(g,A'))$ I need to construct $[E']\in\mathrm{Ext}_R^1(C,A')$, such that $[E_1]=\mathrm{Ext}_R^1(g,A')([E'])=[E'g]$. This equivalent to existence of morphism of extensions $\Gamma:E_1\to E'$ of the form $(1_{A'}, \beta,g)$, for some $R$-homomorphism $\beta$.
$$
\begin{array}{cccccccccc}
&&&&&&&0&&&\\
&&&&&&&\downarrow &&&\\
E_1f: & 0 & \xrightarrow{} & A' & \xrightarrow{f_f} & B_f & \xrightarrow{g_f} & A & \xrightarrow{} & 0 \\
&&&  \downarrow 1_A' && \downarrow \beta_f && \downarrow f &&&\\
E_1: & 0 & \xrightarrow{} & A' & \xrightarrow{f_1} & B_1 & \xrightarrow{g_1} & B & \xrightarrow{} & 0 \\
&&&  \downarrow 1_A' && \downarrow ? && \downarrow g &&&\\
E': & 0 & \xrightarrow{} & A' & \xrightarrow{?} & ? & \xrightarrow{?} & C & \xrightarrow{} & 0 \\
&&&&&&&\downarrow &&&\\
&&&&&&&0&&&\\
\end{array}
$$
Question. How should I define $E'$, and how to use here that $[E_1f]$ splits?

REPLY [3 votes]: I don't want to spoil this for you so I will only give a hint, let me know if you would like more.  Life becomes easier if you make some identifications: the top row middle term may as well be $A \oplus A'$ because of the splitting, and let's identify $A$ with a submodule of $B$ via $f$, and replace $C$ with $B/A$. Now $\beta_f$  is a map $A' \oplus A \to B_1$.  My hint is that the missing middle term should be $B_1/\beta_f(A)$ and the map $B_1/\beta_f(A) \to B/A$ is $b + \beta_f(A) \mapsto g_1(b)+A$ (you need to check it is well-defined...).