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:

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

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

Since both $\mathrm{Ext}^1(B,A)$ and $\mathrm{Hom}(A,E)$ are groups, I wonder if there is more generally a homomorphism from the cyclic subgroup $\langle \eta\rangle \le \mathrm{Hom}(A,E)$, mapping an element $n\cdot \eta$ to $n\cdot E$, where the multiplication on the right is taken with respect to the Baer sum.

Thanks in advance, a reference would be highly appreciated.

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.