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In case it's necessary, I'm working in the category $\mathbf{Ab}$ of abelian groups. My question concerns what I find to be a strange way of viewing the elements of the Ext group $\mbox{Ext}(A,B)$ of the abelian groups $A$ and $B$. The statement of my question is at the end of this post but first two definitions and some notation.

A morphism $\varphi\colon E\rightarrow E'$ of short exact sequences $$E=0\rightarrow B\rightarrow G\rightarrow A\rightarrow 0$$ and $$E'=0\rightarrow B'\rightarrow G'\rightarrow A'\rightarrow 0$$ of abelian groups is a triple $\varphi=(b,g,a)$ of group homomorphisms $b\colon B\rightarrow B'$, $g\colon G\rightarrow G'$ and $a\colon A\rightarrow A'$ making the obvious diagram commute. So, $\varphi$ is an isomorphism precisely when $b, g$ and $a$ are isomorphisms.

We say that the SESs (short exact sequences) $E$ and $E'$ are equivalent if $B=B'$, $A=A'$ and there exists a SES morphism $(b,g,a)=\varphi\colon E\rightarrow E'$ with $b=\mbox{Id}_B$ and $a=\mbox{Id}_A$. By the short five-lemma, if $\varphi$ is an equivalence of SESs, then it is a SES isomorphism.

Wikipedia says that the classes of equivalent SESs with fixed $A$ and $B$ are in one to one correspondence with the elements of the Ext group $\mbox{Ext}(A,B)$. However, one would have guessed that it would be more natural to view the elements of $\mbox{Ext}(A,B)$ as classes of isomorphic SESs with $A$ and $B$ fixed. My question is, are these actually equivalent ways of viewing the elements of the Ext group?

More precisely, if $E$ and $E'$ are short exact sequences of abelian groups with $A=A'$ and $B=B'$, is it true that if $E$ is isomorphic to $E'$ then $E$ is equivalent to $E'$? (If this is true, does it then hold in any exact category?)

Dan Rust
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    Just as a remark, two isomorphic short exact sequences will be equivalent iff there is an automorphism of $G$ that simultaneously extends $b$ and lifts $a$. It seems unlikely that this can be done in general but I don't have a counterexample. – Miha Habič Apr 04 '13 at 23:47

2 Answers2

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The answer is no.

Consider a prime number $p>2$ and set $$ E: \quad 0 \longrightarrow \mathbb{Z} \overset{\cdot p}\longrightarrow \mathbb{Z} \overset{\pi}\longrightarrow \mathbb{Z}/p \longrightarrow 0 $$

$$ E': \quad 0 \longrightarrow \mathbb{Z} \overset{\cdot p}\longrightarrow \mathbb{Z} \overset{\cdot 2 \circ \pi}\longrightarrow \mathbb{Z}/p \longrightarrow 0. $$ Here the map $\pi$ is the standard projection, and $\cdot 2 \circ \pi$ is the canonical projection followed by the group automorphism of multiplication by two on $\mathbb{Z}/p$.

There is an obvious isomorphism of SES's, however one can check there cannot exist an equivalence: let's use your notation for maps $b,g,a$ and suppose it is an equivalence. Since $g$ must be an isomorphism, it is either multiplication by -1 or the identity. However, since $b= id_{\mathbb{Z}}$, we find $g=id_{\mathbb{Z}}$ as well. But then we run into problems since the maps on the right will not commute.

There are way more extensions of $\mathbb{Z}/p$ by $\mathbb{Z}$ up to equivalence then there are up to isomorphism, i believe to remember that there are $p$ of the first kind (the ones as above with $2$ being any number between 1 and $p-1$, plus the split extension) and only two of the second kind, but these numbers might be incorrect.

Please let me know if this was the answer you were looking for!

First original answer, incorrect: Counterexamples are not hard to find. For example take two copies of the split exact sequence $$ 0 \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow 0 $$ and all maps $b,g,a$ multiplication by $-1$. Clearly $A=A'$ and $B=B'$ and it is an isomorphism of SES, but not an equivalence.

In my homological algebra class, i had a similar remark; the teacher then decided to say ultraisomorphism for what you call equivalence. (He is quite funny and a great teacher, but that aside)

The philosophy is the following: Such a short exact sequence is called an extension of $C$ bt $A$ (or the other way around, everybody always forgets) and one is interested in the possible ways to build such an extension. Seeing the groups $A$ and $C$ as your fixed building blocks, it comes as no surprise that you do not want to change them in any way, not even by an isomorphism.

Joachim
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  • @Joachim: Thanks for the answer. As Miha says, this isn't quite what I'm looking for. A counter example would constitute two isomorphic SESs with $A=A'$, $B=B'$ but for which no equivalence exists. – Dan Rust Apr 05 '13 at 00:25
  • Ah, i see.. I read "isomorphism" instead of isomorphic. I'll think about it. – Joachim Apr 05 '13 at 11:29
  • Nitpick: you want $p > 3$, because you need $2 \neq -1$. –  Apr 05 '13 at 12:29
  • Fixed, thanks @Hurkyl! – Joachim Apr 05 '13 at 13:48
  • Thanks. Your new answer is really simple. – Dan Rust Apr 05 '13 at 13:55
  • @Hurkyl, i just realized that in fact the argument still holds even if $p=3$. Because the middle map is forced to be the identity anyway! Or am i wrong? – Joachim Apr 05 '13 at 13:56
  • @DanielRust, you're welcome. It was actually given in the class i already mentioned, so it might be some sort of standard example.. Bet regards, Joachim. – Joachim Apr 05 '13 at 13:57
  • ah right. I didn't look at the rest of the diagram when I thought about that –  Apr 05 '13 at 14:09
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I've been thinking about this same issue lately, and one thing that has clarified it is remembering that $\operatorname{Ext}(\bullet, \bullet)$ is a bifunctor, like $\operatorname{Hom}(\bullet, \bullet)$. So if we had a map between short exact sequences which was the identity on $B$ but not on $A$, then that non-identity map (call it $\varphi$) on $A$ should in general induce a non-identity endomorphism $ \operatorname{Ext}(\varphi,B)$ on $\operatorname{Ext}(A,B)$.

Eric Auld
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