Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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...).

share|cite|improve this answer
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
For diagram-chase problems like this I find it helps to write things explicitly and have as little notation as possible, yes. – m_t_ May 11 '12 at 15:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.