# How to prove $\mathrm{Im}(\mathrm{Ext}_R^1(g,A'))=\mathrm{Ker}(\mathrm{Ext}_R^1(f,A'))$

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?

-
This is heroic use of the array environment. – Dylan Moreland May 11 '12 at 0:52
That's why I got upvote for the question? – Norbert May 11 '12 at 0:53
No, I upvoted it because I like homological algebra and you showed a lot of your work even though it was difficult to do so. I'm not (entirely) swayed by pretty diagrams. – Dylan Moreland May 11 '12 at 0:54
Thanks, @DylanMoreland :) Homological algebra blows my mind! – Norbert May 11 '12 at 0:57

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...).
Thanks for your reply. I've checked all the conditions. This indeed gives exact sequence $E'$ and morphism of extinsions. Could you explain what ideas do you have used. I can perform computations and basic manipulations in homological algebra but I have no intuition here. – Norbert May 11 '12 at 13:07
@Norbert When I first looked at your problem I couldn't see what the answer was, but when I wrote it down stripping out as much of the notation as possible it was a lot clearer. ($C$, $f$, $g$,...can all be suppressed without changing the problem). When you do this you see that the missing module ought to be something that is "$\dim A$ smaller" than $B_1$, and it should be missing the bit of $B_1$ that maps to $A\subset B$. Then it is natural to try what I said. – m_t_ May 11 '12 at 13:48
So the idea is to use as much explicit constructions as possible. Like $C\simeq B/A$ – Norbert May 11 '12 at 14:00