This question was asked by a student (in a slightly different form), and I was unable to answer it properly. I think it's quite interesting.


The problem is to produce an example of the following situation: find a short exact sequence
$$ 0 \to X\_1 \to X\_2 \to X\_3 \to 0$$
(in some category of your choice), and a second exact sequence
$$ 0 \to Y\_1 \to Y\_2 \to Y\_3 \to 0$$
such that there are isomorphisms $X\_n \cong Y\_n$ for all $n$, BUT in such a way that there is no commutative diagram whatsoever between the two sequences, with the vertical maps being isomorphisms.


It is impossible to find such an example in the category of vector spaces, or of finitely-generated abelian groups. I don't know about the general case, though.


I would be grateful for any example, but would be disappointed if the chosen category were constructed specifically to answer the problem.


EDIT / COMMENT: in the first version of this question I was asking for sequences which could potentially be infinite. Some great examples came in the comments straightaway. I'm very thankful for them, but I recall only now that the student's original question was about short exact sequences, so I've edited accordingly. (I'm sorry for the confusion, the student asked me this question several months ago, and I was posting only now, for some reason... and got it wrong. I'm very happy to know about the examples involving infinite sequences, though.)



(Short) Exact sequences with no commutative diagram between them
$\def\ZZ{\mathbb{Z}}$ 
This can happen in finitely generated abelian groups. Let $p$ be prime, set $G = \ZZ/p^2 \oplus \ZZ/p$ and set $H = \ZZ/p^3 \oplus \ZZ/p^2 \oplus \ZZ/p$. Then there are two non-isomorphic short exact sequences $0 \to G \to H \to G \to 0$. The first one is the sum of the extensions:
$$\begin{matrix}
0 & \to & \ZZ/p & \to & \ZZ/p^3 & \to & \ZZ/p^2 &\to& 0 \\
0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 & \to & 0 &\to& 0 \\
0 & \to & 0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 \\
\end{matrix}$$
and the second is the sum of
$$\begin{matrix}
0 & \to & \ZZ/p^2 & \to & \ZZ/p^3 & \to & \ZZ/p &\to& 0 \\
0 & \to & 0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 &\to& 0 \\
0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 & \to & 0 \\
\end{matrix}$$


Let's see that these are not isomorphic. Write $\alpha$ for the map $G \to H$. In the first extension, $\alpha(p G) \cap p^2 H = (0)$; in the second extension, $\alpha(pG) = p^2 H$. 


In any reasonable category, isomorphism classes of extensions $0 \to X \to ?? \to Z \to 0$ are classified by the orbits of $\mathrm{Aut}(X) \times \mathrm{Aut}(Z)$ on $\mathrm{Ext}^1(Z,X)$. You are asking for cases where there is more than one orbit which makes the center term isomorphic as an abstract element of the category. There is no reason this shouldn't happen, so I would expect it to happen basically any time there are nontrivial extension groups available.




---


To add some more high level context, the analogous example works in $k[t]$-modules. (That is to say, extending $k[t]/t^2 \oplus k$ by itself to get $k[t]/t^3 \oplus k[t]/t^2 \oplus k$.) In general, for any partition $\lambda$, let $M(\lambda)$ be the $k[t]$ module $\bigoplus\_i k[t]/t^{\lambda\_i}$. Then $\mathrm{Ext}^1(M(\mu), M(\lambda))$ is stratified into locally closed pieces according to the isomorphism type of the extension, and the number of components where the extension is isomorphic to $M(\nu)$ is the Littlewood-Richardson number $c\_{\lambda \mu}^{\nu}$. This example is $c\_{(21) (21)}^{(321)}=2$. Since $Aut(M(\lambda))$ is always a connected group (it is a unipotent extension of $\prod GL(\lambda\_i)$), it can't mix the components.


You can start to get context for this from the first two chapters of Schiffmann's [Lectures on Hall Alegbras](http://arxiv.org/abs/math/0611617). Unfortunately, Schiffmann always takes $k$ to be a finite field so that he can count points, which makes it harder to talk about positive dimensional subvarieties of $\mathrm{Ext}^1$.


