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Let $\mathsf A$ be an abelian category and $\mathsf{K(A)}$ be the homotopy category of chain complexes over $\mathsf A$. Let $P_\bullet,Q_\bullet$ be projective resolutions of $A,B\in \mathsf A$ respectively. The fundamental lemma of homological algebra says $\mathsf{Hom}_\mathsf{A}(A,B)\cong [P_\bullet,Q_\bullet]$. This means that the set map taking an object of $\mathsf A$ to the homotopy class of its projective resolutions extends to a functor $P:\mathsf A\rightarrow \mathsf{K(A)}$. (See page 16 these notes.)

By homotopy invariance, chain homology lifts to $\mathsf{K(A)}$. In particular, we have a functor $H_0:\mathsf{K(A)}\rightarrow \mathsf A$.

Is $P \dashv H_0$?

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No. For example, let $X$ and $Y$ be two objects of $\mathsf{A}$, and denote by $Y[n]$ the complex with $Y$ in degree $n$ and zero in other degrees, so that if $n\neq0$ then $H^0(Y[n])=0$.

Then for $n>0$, $\operatorname{Hom}_{\mathsf{A}}\left(X,H^0(Y[n])\right)=0$, but $\operatorname{Hom}_{\mathsf{K(A)}}\left(P(X),Y[n]\right)=\operatorname{Ext}^n(X,Y)$, which is in general not zero.

[Something related that is true: $P$ extends to a functor $\mathsf{D(A)}\to\mathsf{K(A)}$ from the derived category of $\mathsf{A}$. This is left adjoint to the natural functor $\mathsf{K(A)}\to\mathsf{D(A)}$.]

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