A chain map is a quasi-isomorphism exactly when its cone is acyclic

Assume $\mathrm{Cone}(f)$ is acyclic. If $[x]\in H_{n-1}(C)$ satisfies $H_{n-1}(f)([x])=0$, choose $y\in D_n$ with $d_n^D(y)=f_{n-1}(x)$. Then $(-y,x)$ is an $n$-cycle of the cone by [L2], so [L3] gives $(-y,x)=d_{n+1}^{\mathrm{Cone}(f)}(z,w)$ for some $(z,w)$. Hence $x=-d_n^C(w)$, so $H_{n-1}(f)$ is injective. Likewise, if $y\in Z_n(D)$, then $(y,0)$ is an $n$-cycle of the cone, so acyclicity gives $(y,0)=d_{n+1}^{\mathrm{Cone}(f)}(z,w)$. Thus $y=d_{n+1}^D(z)+f_n(w)$, and $H_n(f)$ is surjective.

Conversely, assume $f$ is a quasi-isomorphism. Let $(y,x)$ be an $n$-cycle of $\mathrm{Cone}(f)$. Then $x\in Z_{n-1}(C)$ and $f_{n-1}(x)=-d_n^D(y)$ is a boundary in $D$, so [L1] gives $x=d_n^C(w)$ for some $w\in C_n$. Then $y+f_n(w)$ is an $n$-cycle in $D$, so [L1] again gives $z\in Z_n(C)$ and $v\in D_{n+1}$ with $y+f_n(w)=f_n(z)+d_{n+1}^D(v)\mathrm{.}$ Therefore $(y,x)=d_{n+1}^{\mathrm{Cone}(f)}(v,z-w),$ so every cone cycle is a boundary. By [L3], $\mathrm{Cone}(f)$ is acyclic, and together with [L1] this proves the equivalence.