# $Ext^n(A,B)\cong H^n(Hom(A,I^\bullet))$, with $I^\bullet$ an injective resolution of B

Weibel 2.5.1 Equivalent statements of injective $R$-module.

but he answer uses a fact I am not familiar with:

Fact 1 If $$B\to I^0\to I^1\to \cdots \to I^n\to\cdots$$ is any injective resolution of $$B$$, then for any $$A$$, $$Ext^n(A,B)\cong H^n(Hom(A,I^\bullet))$$

Could you prove it or point to a resource where it is proven?

• To me, this is the definition of $\mathrm{Ext}$. I'd guess you're using the symmetric one (derived functor of $\mathrm{Hom}(A,-)$ instead of $\mathrm{Hom}(-,B)$) and haven't established symmetry yet, but the question is a bit odd without specifying. Commented Nov 28, 2023 at 20:43

Though the fact about being able to compute $$\mathsf{Ext}$$ in two different ways is correct and essential, it is not necessary for the purposes of understanding (in the notation of the linked post) why $$(1)\implies(3)$$. I personally think it is more straightforward to show $$(1)\iff(2)\implies(3)\implies(4)\implies(1)$$.
If you a priori define $$\mathsf{Ext}_R^\ast(A,B)$$ to be the evaluation of the right derived functors of $$\mathsf{Hom}_R(-,B)$$ (with variance $$\mathsf{RMod}^{\mathsf{op}}\to\mathsf{Ab})$$ at $$A$$, then injectivity of $$B$$ implies exactness of this functor hence all its derived functors vanish in positive degree.
If instead you a priori (the two definitions are the same up to canonical isomorphism) define $$\mathsf{Ext}_R^\ast(A,B)$$ to be the evaluation of the right derived functors of $$\mathsf{Hom}_R(A,-)$$ at $$B$$, then "Fact 1" is true by definition/by explicit construction and there is really nothing to say; it is immediate.
And either way, you should define $$\mathsf{Ext}_R^\ast$$ as one of the above... and both roads lead to Rome $$((1)\implies(3))$$. If I rememeber correctly, Weibel uses the first of the two definitions. This is easier. It is true and very importantly true that we can "balance" $$\mathsf{Ext}_R^\ast$$ to obtain isomorphisms between both roads, which is shown by Weibel later on in the chapter, but we don't actually need that.