Ext functor isomorphism

Question

I am having a hard moment trying to solve something that looks really easy but I cannot solve. I want to prove that, given two presentations of left R-modules, one injective and the other projective, respectively, $0\to A \to I \to B \to 0$ and $0\to C \to P \to D \to 0,$ then we have isomorphisms $Ext_{R}^{i}(C, A) \cong Ext_{R}^{i}(D,B) $ for all $i>0.$ I have the feeling that using the long exact sequence derived from the short ones the cases where $i>1$ are easy. But I am having a lot of troubles with the case $i=1$ because $Hom(C, I)$ and $Hom(P, B)$ are not necessarily 0.

Answer 1

For all $R$-modules $X$, for all $k\ge 1$, one has an isomorphism $$\text{Ext}^k(X,B)\xrightarrow{(1)} \text{Ext}^{k+1}(X,A), $$ because $\text{Ext}^k(X,I) = \text{Ext}^{k+1}(X,I)= 0$.

For all $R$-modules $Y$, for all $k\ge 1$, one has an isomorphism $$\text{Ext}^k(C,Y) \xrightarrow{(2)}\text{Ext}^{k+1}(D,Y),$$ because $ \text{Ext}^{k}(P,Y)= \text{Ext}^{k+1}(P,Y)=0$.

Putting these together, (for $k\ge 1$) one has an isomorphism $$\text{Ext}^k(D,B)\xrightarrow{(1)} \text{Ext}^{k+1}(D,A)\xleftarrow{(2)} \text{Ext}^k(C,A).$$