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?