Proposition 1

If C is zero on the right and C' zero on the left, then HomgrA (C, C') is zero on the left, and the canonical k-linear mapping*

is bijective.

We have exact sequences

On the other hand $\mathrm{Homgr}_A^0 (C, C')$ is identified with $\mathrm{Hom}_A (C_0, {C'}^0)$, $Z^0(\mathrm{Homgr}_A (C, C'))$ then being identified with the set of $f : C_0 \to {C'}^0$ such that ${d'}^0 \circ f = 0, f \circ d_1 = 0$; $B^0(\mathrm{Homgr}_A (C, C'))$ is zero; finally the mapping $\lambda^0$ associates to the class of $f$ modulo $\{0\}$ the homomorphism $\varphi : H_0(C) \to H^0(C')$ such that $f = i \circ \varphi \circ p$, whence the proposition.

Let $u : \tilde{C} \to C$ and $u' : C' \to \tilde{C}'$ be morphisms of complexes; then the canonical homomorphism $\mathrm{Homgr}_A (u, u') : \mathrm{Homgr}_A (C, C') \to \mathrm{Homgr}_A (\tilde{C}, \tilde{C}')$, defined by $f \mapsto u' \circ f \circ u$, is a morphism of complexes, as follows immediately from formula (1). Moreover, the following diagram is commutative