# If $\mathrm{Hom}_R(f,N)=\mathrm{Hom}_R(M,g)$, is it true that $\mathrm{Ext}_R^i(f,N)=\mathrm{Ext}_R^i(M,g)$?

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.

• 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. Apr 20, 2021 at 23:26

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 $$g=\text{id}_N$$.