[proofplan]
The mapping cone fits into the standard [long exact homology sequence](/theorems/4210) relating $H_n(C)$, $H_n(D)$, and $H_n(\operatorname{Cone}(f))$. If the cone is acyclic, the long exact sequence collapses around each degree and forces $H_n(f)$ to be an isomorphism. Conversely, if each $H_n(f)$ is an isomorphism, exactness shows that every map into and out of $H_n(\operatorname{Cone}(f))$ has the only possible image and kernel, so $H_n(\operatorname{Cone}(f))$ must be zero.
[/proofplan]
[step:Use the long exact homology sequence of the mapping cone]
For the chain map $f: C \to D$, define the mapping cone chain complex $\operatorname{Cone}(f)$ by
\begin{align*}
\operatorname{Cone}(f)_n &= D_n \oplus C_{n-1},
\end{align*}
with its standard cone differential. The cone construction gives a short exact sequence of chain complexes
\begin{align*}
0 \longrightarrow D \longrightarrow \operatorname{Cone}(f) \longrightarrow C[-1] \longrightarrow 0,
\end{align*}
where $C[-1]$ denotes the shifted complex with $(C[-1])_n = C_{n-1}$.
Applying the long exact homology sequence associated to a short exact sequence of chain complexes (citing a result not yet in the wiki: long exact sequence in homology associated to a short exact sequence of chain complexes), we obtain, for every $n \in \mathbb Z$, an exact segment
\begin{align*}
H_n(C) \xrightarrow{H_n(f)} H_n(D)
\longrightarrow H_n(\operatorname{Cone}(f))
\longrightarrow H_{n-1}(C)
\xrightarrow{H_{n-1}(f)} H_{n-1}(D).
\end{align*}
Here $H_n(f): H_n(C) \to H_n(D)$ is the map induced by the chain map $f$ on degree-$n$ homology.
[/step]
[step:Deduce that acyclicity of the cone makes each homology map an isomorphism]
Assume that $\operatorname{Cone}(f)$ is acyclic. Then for every $n \in \mathbb Z$,
\begin{align*}
H_n(\operatorname{Cone}(f)) = 0.
\end{align*}
The exact segment from the previous step becomes
\begin{align*}
H_n(C) \xrightarrow{H_n(f)} H_n(D)
\longrightarrow 0
\longrightarrow H_{n-1}(C)
\xrightarrow{H_{n-1}(f)} H_{n-1}(D).
\end{align*}
Exactness at $H_n(D)$ gives
\begin{align*}
\operatorname{im} H_n(f) = \ker\bigl(H_n(D) \to 0\bigr) = H_n(D),
\end{align*}
so $H_n(f)$ is an epimorphism. Exactness at $H_n(C)$, equivalently after shifting the displayed exact segment by one degree, gives
\begin{align*}
\ker H_n(f) = \operatorname{im}\bigl(0 \to H_n(C)\bigr) = 0,
\end{align*}
so $H_n(f)$ is a monomorphism. Since $\mathcal A$ is abelian, a morphism that is both monic and epic in this exact-sequence sense is an isomorphism. Therefore $H_n(f)$ is an isomorphism for every $n \in \mathbb Z$, and hence $f$ is a quasi-isomorphism.
[/step]
[step:Deduce that a quasi-isomorphism has acyclic cone]
Assume conversely that $f$ is a quasi-isomorphism. Thus, for every $n \in \mathbb Z$, both morphisms
\begin{align*}
H_n(f): H_n(C) &\to H_n(D), \\
H_{n-1}(f): H_{n-1}(C) &\to H_{n-1}(D)
\end{align*}
are isomorphisms.
Fix $n \in \mathbb Z$. In the exact segment
\begin{align*}
H_n(C) \xrightarrow{H_n(f)} H_n(D)
\longrightarrow H_n(\operatorname{Cone}(f))
\longrightarrow H_{n-1}(C)
\xrightarrow{H_{n-1}(f)} H_{n-1}(D),
\end{align*}
the map $H_n(f)$ is an epimorphism, so exactness at $H_n(D)$ gives
\begin{align*}
\ker\bigl(H_n(D) \to H_n(\operatorname{Cone}(f))\bigr)
=
\operatorname{im} H_n(f)
=
H_n(D).
\end{align*}
Hence the morphism $H_n(D) \to H_n(\operatorname{Cone}(f))$ is the zero morphism.
Exactness at $H_n(\operatorname{Cone}(f))$ then gives
\begin{align*}
\ker\bigl(H_n(\operatorname{Cone}(f)) \to H_{n-1}(C)\bigr)
=
\operatorname{im}\bigl(H_n(D) \to H_n(\operatorname{Cone}(f))\bigr)
=
0.
\end{align*}
Thus the connecting morphism $H_n(\operatorname{Cone}(f)) \to H_{n-1}(C)$ is a monomorphism.
Since $H_{n-1}(f)$ is a monomorphism, its kernel is zero. Exactness at $H_{n-1}(C)$ gives
\begin{align*}
\operatorname{im}\bigl(H_n(\operatorname{Cone}(f)) \to H_{n-1}(C)\bigr)
=
\ker H_{n-1}(f)
=
0.
\end{align*}
The connecting morphism is both monic and has zero image, so its domain is the zero object:
\begin{align*}
H_n(\operatorname{Cone}(f)) = 0.
\end{align*}
Because $n \in \mathbb Z$ was arbitrary, $\operatorname{Cone}(f)$ is acyclic.
[/step]
[step:Combine the two implications]
We have shown that acyclicity of $\operatorname{Cone}(f)$ implies that $H_n(f)$ is an isomorphism for every $n \in \mathbb Z$, so $f$ is a quasi-isomorphism. We have also shown that if $f$ is a quasi-isomorphism, then $H_n(\operatorname{Cone}(f)) = 0$ for every $n \in \mathbb Z$, so $\operatorname{Cone}(f)$ is acyclic. Therefore $f$ is a quasi-isomorphism if and only if $\operatorname{Cone}(f)$ is acyclic.
[/step]
