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

Question

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.

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 $g=\text{id}_N$.