Corollary

Let $C, C', D, D'$ be four complexes, $f : C \to C'$ a morphism, $u : D \to C$ and $v : C' \to D'$ homotopy equivalences. For $v \circ f \circ u$ to be a homotopy equivalence (resp. a homology isomorphism), it is necessary and sufficient that $f$ be one.

If $f$ is a homotopy equivalence (resp. a homology isomorphism), then $v \circ f \circ u$ is a composition of homotopy equivalences (resp. homology isomorphisms), hence is one. Conversely, let $\overline{u}$ and $\overline{v}$ be morphisms inverse to $u$ and $v$ up to homotopy; then $\overline{v} \circ (v \circ f \circ u) \circ \overline{u}$ is homotopic to $f$ by prop. 4; whence the conclusion by prop. 5, and the corollary of prop. 3.

One says that the complex $C$ is homotopic to zero if $1_C$ is homotopic to the zero mapping, that is, if there exists a graded $A$-endomorphism $s$ of degree 1 of $C$ such that $1_C = s \circ d + d \circ s$. This also amounts to saying that the unique morphism $0 \to C$ (resp. $C \to 0$) is a homotopy equivalence. A complex homotopic to zero has zero homology (prop. 5).