On equivalent definitions of Ext

Solution 1:

$\DeclareMathOperator{Hom}{Hom}$The main point is that a projective resolution $\dots \to P^{-2} \to P^{-1} \to P^{0} \xrightarrow{\alpha^0} X$ gives you a quasi-isomorphism $\alpha \colon P^{\bullet} \to X[0]$. Indeed, the mapping cone of $\alpha$ is exact because it is the resolution (maybe up to an immaterial sign in the differentials).

A quasi-isomorphism becomes an isomorphism in the derived category (because that's what we invert), in particular precomposition with $\alpha^{-1}$ (or, if you prefer: composition with the roof $X[0] \xleftarrow{\alpha} P^\bullet \xrightarrow{1} P^\bullet$) gives an isomorphism $$\Hom\nolimits_{D(A)}(P^\bullet,Y) \xrightarrow{\cong} \Hom\nolimits_{D(A)}(X[0],Y) $$ for every complex $Y \in D(A)$.

From fact 2) of the question we have a composite isomorphism $$ \Hom\nolimits_{K(A)}(P^\bullet,Y) \xrightarrow{\cong} \Hom\nolimits_{D(A)}(P^\bullet,Y) \xrightarrow{\cong} \Hom\nolimits_{D(A)}(X[0],Y) $$ for all complexes $Y$ which is explicitly given by sending a (homotopy class of a) chain map $f\colon P^\bullet \to Y$ to the roof $X[0] \xleftarrow{\alpha} P^\bullet \xrightarrow{f} Y$, so it only remains to identify the complex $\Hom_{K(\mathscr{A})}(P^\bullet, Y)$.

I hope I got the signs and indices right in what is to follow: for any two complexes $A$ and $B$ over an additive category $\mathscr{A}$ one defines the total $\Hom$-complex $\Hom\nolimits^\bullet(A,B)$ of abelian groups by $$ \Hom\nolimits^k(A,B) = \prod_{n \in \mathbb{Z}}\Hom\nolimits_{\mathscr{A}}(A^n,B^{n+k}) $$ with differential $(f^n)_{n \in \mathbb{Z}} \mapsto \left(f^{n+1}d_A^n-(-1)^{k}d_B^{n+k}f^n \right)_{n\in\mathbb{Z}}$ and it is elementary to check (if indeed I got the signs right) that $$\boxed{ H^k\left(\Hom\nolimits^\bullet(A,B)\right) = \Hom\nolimits_{K(\mathscr{A})}(A,B[k]) }$$ because in order for $f = (f_n)_{n \in \mathbb{Z}}$ to be a cycle in $\Hom\nolimits^k(A,B)$ it is necessary and sufficient that $f \colon A \to B[k]$ defines a chain map and to be a boundary means that it is homotopic to zero. (In the sign conventions I'm used to shifting a complex by $1$ involves multiplying its differentials by $-1$)

If we take $A = P^\bullet$ and $B=Y[0]$ and $k\in\mathbb{Z}$ we see that the $\Hom$ complex collapses to $\Hom^k(P^{\bullet},Y[0]) = \Hom\nolimits_{\mathscr{A}}(P^{-k},Y)$ and computing its cohomology amounts to taking the cohomology of the complex $$ \dots \to 0 \to \Hom\nolimits_{\mathscr{A}}(P^0,Y) \xrightarrow{d^\ast} \Hom\nolimits_{\mathscr{A}}(P^{-1},Y) \xrightarrow{d^\ast} \Hom\nolimits_{\mathscr{A}}(P^{-2},Y) \xrightarrow{d^\ast} \cdots $$ which via $$ \operatorname{Ext}^k(X,Y) = \operatorname{Hom}_{D(A)}(X,Y[k]) \cong \operatorname{Hom}_{K(A)}(P^\bullet,Y[k]) = H^k(\operatorname{Hom}^\bullet(P^\bullet,Y[0])) $$ gives the identification you ask about.