Can two elements of an Ext group come from the same middle object of an SES

abelian-categoriesabstract-algebraderived-functorsexact-sequencehomological-algebra

Let $X$ be an object of an abelian category. Is it possible for there to be an object $B$ that is a subobject of $X$ in two distinct ways that yield isomorphic cokernels but is not off by an automorphism of $X$? More precisely, can we have

$$
B \overset{i}{\hookrightarrow} X \overset{\pi_i}{\twoheadrightarrow} \mathrm{Coker}(i)
\qquad\text{&}\qquad
B \overset{j}{\hookrightarrow} X \overset{\pi_j}{\twoheadrightarrow} \mathrm{Coker}(j)$$

with $\mathrm{Coker}(i) \cong \mathrm{Coker}(j)$, but such that there is no $\phi \in \mathrm{Aut}(X)$ for which $\phi i = j$ and $\pi_i = \pi_j\phi$? In other language, can we have distinct elements of $\mathrm{Ext}^1(A,B)$ that come from the same middle term $X$ of a short exact sequence?

Can such a thing happen outside of an abelian category?

Best Answer

Yes. For instance, in the category of abelian groups, note that if $p$ is a prime, then $\operatorname{Ext}^1(\mathbb{Z}/(p),\mathbb{Z}/(p))\cong\mathbb{Z}/(p)$ has $p$ different elements, but up to isomorphism there are only two groups that can form such an extension, namely $\mathbb{Z}/(p)\oplus\mathbb{Z}/(p)$ and $\mathbb{Z}/(p^2)$. So, if $p>2$, there must be inequivalent extensions that are isomorphic just as groups.

To see such an example very explicitly, consider the two short exact sequences $$0\to\mathbb{Z}/(p)\stackrel{p}\to\mathbb{Z}/(p^2)\to\mathbb{Z}/(p)\to 0$$ and $$0\to\mathbb{Z}/(p)\stackrel{2p}\to\mathbb{Z}/(p^2)\to\mathbb{Z}/(p)\to 0$$ where in both cases the quotient map is the usual one. There is no automorphism $\phi:\mathbb{Z}/(p^2)\to\mathbb{Z}/(p^2)$ that makes these two extensions equivalent: such a $\phi$ would be multiplication by some integer $m$, such that $mp=2p$ in $\mathbb{Z}/(p^2)$ (to commute with the inclusions from $\mathbb{Z}/(p)$) and $m=1$ in $\mathbb{Z}/(p)$ (to commute with the quotient maps to $\mathbb{Z}/(p)$). This is impossible, since $m$ would have to be both $1$ and $2$ mod $p$.

If you want an example that's not in abelian category, just restrict your category to consist of only the objects and maps involved in the example above.

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