A chain map is a homotopy equivalence exactly when its cone is contractible

Suppose first that $f$ has homotopy inverse $g$. Apply [L5] to the arrow comparison from $f:C_{\bullet }\to D_{\bullet }$ to $1_D:D_{\bullet }\to D_{\bullet }$ using $u=f$, $v=1_D$, the trivial homotopy $1_{D}f\simeq 1_{D}f$, and the given homotopy $fg\simeq 1_D$. This gives a chain-homotopy equivalence $\mathrm{\Phi }:\mathrm{Cone}(f)\to \mathrm{Cone}(1_D)\mathrm{.}$ By [L6], the target cone is contractible; the second sentence of [L2] then shows that $\mathrm{Cone}(f)$ is contractible as well.

Conversely, suppose $\mathrm{Cone}(f)$ is contractible. By [L2] and [L3], choose a degree-one endomorphism $H$ of $\mathrm{Cone}(f)$ with $dH+Hd=1_{\mathrm{Cone}(f)}\mathrm{.}$ Write $H_n(y,x)=(a_n(y)+b_{n-1}(x),g_n(y)+u_{n-1}(x)),$ where $a_n:D_n\to D_{n+1}$, $b_{n-1}:C_{n-1}\to D_{n+1}$, $g_n:D_n\to C_n$, and $u_{n-1}:C_{n-1}\to C_n$. Expanding $dH+Hd=1$ with [L4] and comparing the bottom-left, bottom-right, and top-left components gives $g_{n-1}{d}_n^D=d_n^C{g}_n,g_{n-1}{f}_{n-1}-d_n^C{u}_{n-1}-u_{n-2}{d}_{n-1}^C=1_{C_{n-1}},$ $d_{n+1}^D{a}_n+a_{n-1}{d}_n^D+f_n{g}_n=1_{D_n}\mathrm{.}$ Thus $g$ is a chain map, $u$ is a homotopy $gf\simeq 1_C$, and $-a$ is a homotopy $fg\simeq 1_D$. Therefore $g$ is a homotopy inverse of $f$, so [L1] makes $f$ a chain homotopy equivalence.