# Why can a projective resolution of $A$ be used to calculate $Ext_R^n(A,B)$?

I know the definition of $$Ext^n_R(A,B)$$ as the $$n$$th right derived functor of $$Hom_R(A,-)$$ applied to $$B$$, which should be calculated by taking an injective resolution $$I_\bullet$$ of $$B$$ and taking the cohomology of $$Hom_R(A,I_\bullet)$$.

But I've read several texts that define it by taking a projective resolution $$P_\bullet$$ of $$A$$ and defining $$Ext^n_R(A,B)$$ to instead be the cohomology of $$Hom_R(P_\bullet,B)$$. This has particularly come up with respect to group cohomology, where we use a projective resolution of $$\mathbb{Z}$$ as a trivial $$\mathbb{Z}[G]$$ module to calculate $$Ext^n_{\mathbb{Z}[G]}(\mathbb{Z},M)$$.

My question is, why is this valid? Presumably there is some way to show that the resulting cochain complexes are homotopy equivalent or isomorphic, but I cannot figure out how. Thanks for any help

• two complex are quasi-isomorphism,you can see the concrect proof in Rotman.of course,it is trivial if you learn triangulated category or spectral sequence. – Jian Nov 1 '18 at 0:09

This follows from th fact that if $$\varepsilon: P\to A$$ is a projective resolution and if $$\eta : B\to I$$ is an injective one, then the induced arrows $$\hom(A,I) \stackrel{\hom(\varepsilon,1)}\longleftarrow\hom(P,I)\stackrel{\hom(1,\eta)}\longrightarrow \hom(P,B)$$ are quasi-isomorphisms. Here $$\hom(P,I)$$ is a total product complex. This is proved in Weibel's book, for example, and uses the "acyclic assembly lemma", a very elementary instance of a spectral sequence argument. Thus one can compute $$\rm Ext$$ with either injective or projective resolutions.