Homotopically projective bounded above complex

For cochain complexes put ${\underset{}{\mathrm{Hom}}}^r(P,A)={\prod }_n{\mathrm{Hom}}_{\mathcal{A}}(P^n,A^{n+r})$, with $(du{)}^n=d_A{u}^n-(-1{)}^r{u}^{n+1}{d}_P$. This is the reindexing of The Hom complex of chain complexes. A complex $P$ is homotopically projective, or K-projective, if ${\mathrm{Hom}}_K(P,A[r])=0$ for every acyclic complex $A$ and every integer $r$. Equivalently $\underset{}{\mathrm{Hom}}(P,A)$ is acyclic: the degree-zero Hom/homotopy identification of Hom in the homotopy category is zero-degree homology of the Hom complex, applied after shifting $A$, identifies these groups with its cohomology (a boundary differs only by the invertible sign $(-1{)}^r$).

The bounded-above case additionally requires $P^n=0$ for all sufficiently large $n$, as in Bounded, bounded below, and bounded above complexes. Boundedness is not part of the general K-projective predicate. Nor is termwise projectivity: a contractible complex has zero Hom from it in $K$ and is K-projective irrespective of its terms.