# Why is $\hbox{Ext}_R^* (M,M) = H^*(\hbox{Hom}_R^*(P^*,P^*))$?

Let me first fix some notation and conventions.

Let $R$ be a ring and $M$ a left $R$-module. Given chain complexes $P^*$ and $Q^*$ in $R$-mod, define $\hbox{Hom}^*_R(P^*,Q^*)$ to be the graded space such that $\hbox{Hom}_R ^i (P^*, Q^*) = \bigoplus_n \hbox{Hom}_R(P^n, Q^{n-i})$. $\hbox{Hom}^*_R (P^*,Q^*)$ has a differential with degree $+1$. Now let $P^*$ be a projective resolution of $M$.

Why is $\hbox{Ext}^*_R (M,M)= H^* ( \hbox{Hom}_R^* (P^*, P^*))$?

I've seen this result stated in a few places but I dont see why it's true. I think I could believe that $\hbox{Ext}^*_R (M,M) = H^* ( \hbox{Hom}_R^*(P^*, M))$. But why is there a projective resolution in the second slot of the Total Hom and not an injective resolution?

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@nik, Off the top of my head: Koszul Duality Patterns in Representation Theory by Beilinson, Ginzburg and Soergel. But there are a few others... –  Anette Nov 26 '13 at 18:20
A projective resolution in an abelian category $\mathcal{A}$ is also an injective resolution in $\mathcal{A}^\mathrm{op}$. (Recall that $\mathrm{Hom}$ is contravariant in the first variable...) –  Zhen Lin Nov 26 '13 at 18:32
If im not mistaken if we have projective resolution on the first slot it does not matter what representant of class of quasiisomorphic complexes we take. If we have chain complexes bounded below then projective resolution makes sure on first slot make give as real derived functor? If im wrong pz let me know. –  user52045 Nov 26 '13 at 18:33
@user52045 I agree! –  uncookedfalcon Jan 10 '14 at 20:03

They key here is that the differential of $\hom^\bullet(P_\bullet, Q_\bullet)$ is designed so that $$H^i(\hom^\bullet(P_\bullet, Q_\bullet)) = \hom_{\mathsf{K}(R)}(P_\bullet, Q_\bullet[-i])$$ where $\mathsf{K}(R)$ is the homotopy category (you might want $i$ instead of $-i$, it depends on your indexing conventions and I can't remember what you should get in this case).

If $P_\bullet$ is a projective resolution then it is colocal with respect to acyclic complexes. This is what Daniel Murfet calls a hoprojective resolution (google that term to find his notes on derived categories) and it has the property that $$\hom_{\mathsf{K}(R)}(P_\bullet, Q_\bullet[-i]) = \hom_{\mathsf{D}(R)}(P_\bullet, Q_\bullet[-i])$$ where $\mathsf{D}(R)$ is the derived category. Now if $Q_\bullet = P_\bullet$ then it is quasi-isomorphic to $M$ as a complex ($M$ as a complex has an $M$ in the zeroth position and has $0$'s elsewhere). Quasi-isomorphisms in $\mathsf{D}(R)$ are actually isomorphisms so $$H^i(\hom^\bullet(P_\bullet, P_\bullet)) = \hom_{\mathsf{D}(R)}(P_\bullet, P_\bullet[-i]) = \hom_{\mathsf{D}(R)}(P_\bullet, M[-i]) = H^i(\hom^\bullet(P_\bullet, M))$$ Now just check that $\hom^\bullet(P_\bullet, M)$ when $M$ is considered a complex is the same as $\hom(P_\bullet, M)$ when $M$ is merely a module, and the homology of that last thing becomes $\mathrm{Ext}^i(M, M)$.

Edit: Just fyi, my experience is with dg-algebras in characteristic $2$, so forgive me if I've left out any necessary $\pm1$'s, because they don't matter to me!

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Thanks for your detailed answer Jim! If I understand correctly the main point is that since $P_\bullet$ is quasi isomorphic to $M$, in the derived category they are isomorphic so the hom spaces in the derived category must be the same? Also, I very much like Murfet's notes. –  Anette Nov 26 '13 at 21:06
Yep, that's the gist :) –  Jim Nov 26 '13 at 21:25

I want to justify the answer in user52045's comment by a reference: Brown, Cohomology of Groups, Theorem I.8 (8.5) states:

If $f: C \to D$ is a quasi-isomorphism between arbitrary complexes and $P$ is a non-negative complex of projectives (over an ring $R$), then $Hom^\ast_R(P,f): Hom^\ast_R(P,C)\to Hom^\ast_R(P,D)$ is a quasi-isomorphism.

Now let $M[0]$ be the complex having $M$ in degree 0 and zero elsewhere. Thus a projective resolution $P \to M$ is a quasi-isomorphism $f:P \to M[0]$ and by the theorem above: $$H^\ast Hom^\ast_R(P,P) \cong H^\ast Hom^\ast_R(P,M[0]) = H^\ast Hom_R(P,M)=Ext^\ast_R(M,M)$$

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