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Apologies in advance if this question is too basic, I looked briefly through Weibel and the stacks project and couldn't find any relevant reference.

Let $\mathcal{A}$ denote an abelian category, and let $A,B$ be a pair of objects in $\mathcal{A}$. Let $E\in\mathrm{Ext}^1(B,A)$ fit in the exact sequence:

$$0\to A\xrightarrow{\eta}E\xrightarrow{\pi} B\to 0.$$

It is well known that $E$'s inverse with respect to the Baer sum is the extension: $$0\to A\xrightarrow{-\eta}E\xrightarrow{\pi} B\to 0,$$ where the inverse is taken in $\mathrm{Hom}(A,E)$.

More generally, for every $n$, assuming $A$ has no torsion, the multiplication by $n$ map gives us an automorphism of $A$, and its composition with $\eta$ gives us a different embedding of $A$ in $E$ with isomorphic image, hence, possibly a different extension.

My question is if the multiplication by $n$ on $\eta$ and multiplication by $n$ on $E$ give isomorphic extensions as seems to be suggested for $n=1,0,-1$.

Thanks in advance, a reference would be highly appreciated.

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    $\begingroup$ The question is unclear to me. For $n = 0$, multiplication by $n$ in $\operatorname{Ext}^1(B,A)$ replaces $E$ by the split extension $A \oplus B$, so cannot be obtained in this way. What do you mean by 'isomorphic extensions' — that the new $E'$ you get is abstractly isomorphic to $E$? (Of course you will not get an isomorphic short exact sequence.) Btw, since you know about Baer sum, you certainly know how to compute $n[E]$ in $\operatorname{Ext}^1(B,A)$, right? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 27 at 17:15
  • $\begingroup$ I agree about the peculiarity of the $n=0$ case, but assuming $A$ has no $n$ torsion or that the entire category has no torsion, replacing $\eta$ changes the embedding of $A$ in $E$, hence results in a possibly non-isomorphic extension. $\endgroup$
    – kindasorta
    Commented Nov 27 at 17:50
  • $\begingroup$ As for your question, we should take a direct sum of $n$-fold copies of $E$, which correspond to an extension of the $n$-fold direct sum of $B$ by that of $A$, and then pullback along the $n$-fold diagonal, and quotienting by the kernel of the summation. $\endgroup$
    – kindasorta
    Commented Nov 27 at 17:52
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    $\begingroup$ Ah sorry, I guess I didn't mean the Baer sum itself, but rather functoriality in $A$. That tells you that $n[E]$ is represented by the pushout of $A \to E$ along $n \colon A \to A$. This explains why you get something isomorphic to $E$ if $n$ is invertible and a split extension if $n$ acts by $0$ on $A$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 27 at 20:13
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    $\begingroup$ In the category of abelian groups $\mathbb{Z}$ has no torsion and multiplication by $n$ is not an automorphism. $\endgroup$
    – Fernando Muro
    Commented Nov 27 at 21:19

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Here is my own try for an answer, assuming further that multiplication by $n$ on $\mathcal{A}$ is invertible for all $n\neq 0$, and specializing to a category of modules for simplicity. I wish to show that the Baer sum $nE = E + \ldots + E$ corresponds to replacing $\eta$ by $n\eta$.

My idea is this, let $P$ denote the pullback of $E\oplus\ldots \oplus E$ (the $n$-fold direct sum) by the $n$-fold diagonal $B\to B\oplus\ldots \oplus B$. It is an extension of $B$ by the $n$-fold direct sum of $n$-copies of $A$.

Its elements correspond to $n$-tuples of elements of $E$ which have the same image in $B$. There is an obvious trace map from $P$ to $E_n$, which is defined to be the object $E$ with the extension structure corresponding to replacing $\eta$ by $n\eta$.

Since the trace map has a section (assuming multiplication by $n$ is an isomorphism, as $\mathcal{A}$ is torsion free), the trace map is an epimorphism. Its kernel consists of $n$-tuples $(e_1,\ldots,e_n)$ which sum to $0$ in $E_n$ and have the same image in $B$, denote this image by $b$, then $nb=0$, again under our torsion freeness assumption, $b=0$.

Therefore all the $e_i$ are in $A$ and sum to 0, hence the kernel of the projection is the kernel of the $n$-fold summation on $A$, hence $E_n$ is isomorphic to $nE$, by definition.

We can see that this arguement wouldn't work for $n=0$, which is indeed an exceptional case.

Still unclear what happens if $E$ or $B$ are allowed to have $n$-torsion.

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