Proposition 4

Let $u : C' \to C$ be a morphism of complexes, $P$ a projective complex, $E$ an injective complex.

a) If $P$ is bounded on the right, or if $C$ and $C'$ are bounded on the right, then

is a morphism.

b) If $E$ is bounded on the left, or if $C$ and $C'$ are bounded on the left, then

is a morphism.

Assume first that $u$ is injective and set $C'' = \operatorname{Coker}\ u$. Since $u$ is a morphism, $C''$ is of zero homology. We have, on the other hand, exact sequences (Proposition 2)

==========

By Lemma 1, $\operatorname{Homgr}_A(P, C'')$ is of zero homology in case $a$, $\operatorname{Homgr}_A(C'', E)$ is of zero homology in case $b$, hence the conclusion.

In the general case, there exists (X, p. 38, cor. to Prop. 7) a complex $\tilde{C}'$, which is right-bounded (resp. left-bounded) when $C$ and $C'$ are, an injective morphism $\tilde{u}: C' \to \tilde{C}'$ and a homotopy $\beta: \tilde{C}' \to C$ such that $u = \beta \circ \tilde{u}$. Then $\tilde{u}$ is a morphism (X, p. 34, cor. to Prop. 5); by what was shown above, $\operatorname{Homgr}_A(l_P, \tilde{u})$ (resp. $\operatorname{Homgr}_A(\tilde{u}, l_E)$) is a morphism in case $a$ (resp. $b$). Moreover, $\operatorname{Homgr}_A(l_P, \beta)$ (resp. $\operatorname{Homgr}_A(\beta, l_E)$) is a homotopy (Prop. 3); therefore, $\operatorname{Homgr}_A(l_P, u)$ (resp. $\operatorname{Homgr}_A(u, l_E)$) is the composite of two morphisms, hence is a morphism.