Let $R$ be an associative unital ring, and let $M$ and $N$ be (left) $R$-modules. Define $R$-module maps $f:M\rightarrow M$ and $g:N\rightarrow N$. Assume further that we have: $$ \mathrm{Hom}_R(f,N)=\mathrm{Hom}_R(M,g) $$

as endomorphisms of $\mathrm{Hom}_R(M,N)$.

I'm wondering if it is then true that, for all $i>0$, $$ \mathrm{Ext}_R^i(f,N)=\mathrm{Ext}_R^i(M,g) $$ as endomorphisms of $\mathrm{Ext}_R^i(M,N)$.

If this result is true, I suspect the proof will involve some double complex, which would be tiresome to type up in latex; a proof, a sketch, or a reference would all be welcome.

If this result is not true, I'm hoping it will at least hold for the special case where $f$ and $g$ are multiplication by some $r\in R$, and again a proof or a reference for this particular case would be appreciated.

  • 3
    $\begingroup$ If $R$ and $M$ are left $R$-modules, $\operatorname{Hom}(M,N)$ is only an abelian group -- it does not have a natural left $R$-module structure in general. $\endgroup$ Apr 20, 2021 at 23:26

1 Answer 1


Let $R=\mathbb{Z}$, $M=\mathbb{Z}/2\mathbb{Z}$ and $N=\mathbb{Z}$. Then $\text{Hom}(M,N)=0$, but $\text{Ext}^1(M,N)\neq0$, so you get a counterexample with $f=0$ and $g=\text{id}_N$.


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