Mapping cone

Remarks

The condition $d^2 = 0$ holds: $d^2(c, y) = \left(d^C d^C c,\ -f d^C c + d^D f c + d^D d^D y\right) = (0, 0)$, because $f$ is a chain map. The inclusion $D_\bullet \to \mathrm{cone}(f)$ and the projection $\mathrm{cone}(f) \to C[-1]_\bullet$, where $C[-1]_n = C_{n-1}$ with differential $-d^C$, form a degreewise split short exact sequence of complexes. In an abelian category, its long exact homology sequence has connecting maps equal to $H_{n-1}(f)$ up to sign. So $f$ is a quasi-isomorphism if and only if $\mathrm{cone}(f)$ has zero homology. Sign conventions differ between texts; they give isomorphic complexes.

Example: let $C_\bullet = D_\bullet = \Z$ in degree $0$ and let $f_0$ be multiplication by $2$. Then $\mathrm{cone}(f)_1 = \Z$, $\mathrm{cone}(f)_0 = \Z$, and the differential is $c \mapsto 2c$, so $H_0(\mathrm{cone}(f)) = \Z/2\Z \ne 0$ and $f$ is not a quasi-isomorphism.

Non-example: dropping the sign, $d(c, y) = (d^C c, f c + d^D y)$, does not give a complex in general. For $C_\bullet = D_\bullet$ with $C_1 = C_0 = \Z$, $d_1 = 1_{\Z}$, and $f$ the identity, the element $(c, 0)$ in degree $2$ goes to $(c, c)$ and then to $(0, 2c) \ne 0$.