I was reading this post
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?
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.
