Why does $H_{n}(\operatorname{Hom}_{R}(M,I_{\ast}(N))=H_{n}(\operatorname{Hom}_{R}(P_{\ast}(M),N))$?

Question

I'm trying to show that the two definitions for the $\operatorname{Ext}$ functor are the same whether obtained via injective or projective resolutions. I understand I need to show the following but can't quite figure it out without somewhere saying that some morphisms or kernels are the same when I'm not convinced that they are. $$H_{n}(\operatorname{Hom}_{R}(M,I_{\ast}(N))=H_{n}(\operatorname{Hom}_{R}(P_{\ast}(M),N))$$ Where $I_{\ast}(N)$ is an injective resolution of $N$ and $P_{\ast}(M)$ is a projective resolution on $M$.

Answer 1

See here for instance to get the definition and main results about the spectral sequence associated to a double complex.

Then we have our projective resolution $P_\bullet \to M$, our injectie resolution $N\to I^\bullet$, with which we form a double complex $K^{p,q} = \hom(P_p, I^q)$. One simply has to be a bit careful with the signs, but it's just bookkeeping, I will not enter into too many details here.

The differential in the $p$-direction is simply $d:=\hom(\partial ,I^q)$ where $\partial$ is the differential for $P_\bullet$, and in the $q$-direction $d':=\hom(P_p,\delta)$ where $\delta$ is the differential for $I^\bullet$.

Then we get two spectral sequences, $\{^IE_r\}$ and $\{^{II}E_r\}$ (in the link I gave they were denoted $'E$ and $''E$ respectively) which both converge to the cohomology of the total complex, with possibly different filtrations.

Now let's look at what they are : $^IE_1^{p,q}$ is the cohomology along the second variable, that is $H^q(\hom(P_p, I^\bullet))$. As $P_p$ is projective, $\hom(P_p,-)$ is exact, and so it commutes with cohomology: $^IE_1^{p,q} = \hom(P_p,N)$ if $q=0$, $0$ else.

For similar reasons, invoking the injectivity of $I^p$, we get $^{II}E_1^{p,q} = \hom(M,I^p), q=0$ and $0$ else.

But this is good : our pages are concentrated on one line !

We therefore easily see that $^IE_2^{p,q} = \mathrm{Ext}^p_{proj}(M,N)$ and $^{II}E_2^{p,q} = \mathrm{Ext}^p_{inj}(M,N)$ where I dentoed $\mathrm{Ext}_{inj}$ or $\mathrm{Ext}_{proj}$ the obvious things.

But then for degree reasons, there are no nonzero differentials on this $E_2$ page, or later : so that's also $E_\infty$. We therefore see that the filtration actually does not matter, because by how they are arranged, there is only one piece of the filtration that is nonzero in every degree, and as the filtration is Hausdorff and exhaustive, it follows that both $\mathrm{Ext}$'s are isomorphic to the cohomology of the total complex, in particular they are isomorphic to one another : there is only one $\mathrm{Ext}$. Tada !

(this happens to be one of the examples that convinced me of the power of spectral sequences)