Show that $\{\mathrm {Hom}_R(A, C_n)\}$ forms a chain complex of abelian groups for every $R$-module $A$ and every $R$-module chain complex $C$. Taking $A = Z_n $, show that if $H_n (\mathrm {Hom}_R(Z_n, C))= 0$, then $H_n(C) = 0$. Is the converse true?
At first I reasoned without plugging the condition $A=Z_n $. Since the chain complex $\mathrm {Hom}_R(A, C)$ is defined from $C $, I suppose that the differential $d_n $ of $\mathrm {Hom}_R(A, C)$ is just the homomorphism that sends $\alpha \in \mathrm {Hom}_R(A, C_n)$ to $u_n \circ \alpha \in \mathrm {Hom}_R(A, C_{n+1})$, with $u_n $ the corresponding differential of $C $. However in this case, I have several doubts.
$\mathrm {Ker} (d_n)$ is the subgroup of $\{\mathrm{Hom}_R(A, C_n)\}$ that are sent by $d_n $ in the zero homomorphism of $A $ into $C_{n+1} $: namely, the subgroup $\{\mathrm{Hom}_R(A, Z_n)\}$. From a similar reasoning follows that $\mathrm {Im} (d_{n-1}) =\{\mathrm{Hom}_R(A, B_n)\}$. So $H_n (\mathrm {Hom}_R(A, C))= \{\mathrm{Hom}_R(A, H_n (C))\}$; however if what I said is correct, there would be no reason to take $A=Z_n $ to prove that if $H_n (\mathrm {Hom}_R(A, C))= 0$, then $H_n(C) = 0$. In fact it is evident that this condition would be true for every $A $ and that the converse, namely if $H_n(C) = 0$, then $H_n (\mathrm {Hom}_R(A, C))= 0$, holds too. I think that very likely I'm wrong somewhere, though my conclusions are not really in contradiction with anything. I would like a confirm or a hint to correct, thank you in advance.
The problem is that the inclusion $H_n (\mathrm {Hom}_R(A, C)) \subset \mathrm{Hom}_R(A, H_n (C))$ isn't always an equality. For instance take $R= \mathbb{Z}$, $C_n = \dfrac{\mathbb{Z}}{8\mathbb{Z}}$ for every $n$ and $u_n x = 4x$
$$ \dots \rightarrow \dfrac{\mathbb{Z}}{8\mathbb{Z}}\xrightarrow{.4} \dfrac{\mathbb{Z}}{8\mathbb{Z}} \rightarrow \dots $$ In particular, $B_n = \{0,4\} \pmod 8 \simeq \dfrac{\mathbb{Z}}{2\mathbb{Z}} $ and $Z_n = \{0,2,4,6\} \pmod 8 \simeq \dfrac{\mathbb{Z}}{4\mathbb{Z}}$ hence ${\rm H}_n = \{0,2\} \pmod 4 \simeq \dfrac{\mathbb{Z}}{2\mathbb{Z}}$.
Now take $A= \dfrac{\mathbb{Z}}{2\mathbb{Z}}$. On the one hand $$ \mathrm{Hom}_R(A, H_n (C)) = {\rm Hom}_\mathbb{Z}\left( \dfrac{\mathbb{Z}}{2\mathbb{Z}} ,\dfrac{\mathbb{Z}}{2\mathbb{Z}}\right) \simeq \dfrac{\mathbb{Z}}{2\mathbb{Z}}. $$ On the other hand, $$ {\rm Hom}_\mathbb{Z}\left( \dfrac{\mathbb{Z}}{2\mathbb{Z}}, B_n\right) = {\rm Hom}_\mathbb{Z}\left( \dfrac{\mathbb{Z}}{2\mathbb{Z}}, Z_n\right) = \left\{ (1 \mapsto 0), (1 \mapsto 4)\right\} $$ since for every $f \in {\rm Hom}_\mathbb{Z}\left( \dfrac{\mathbb{Z}}{2\mathbb{Z}} ,M\right)$ the order of $f(1)\in M$ must divide $2$. Therefore $$H_n (\mathrm {Hom}_R(A, C)) = \{0\}$$ The general principle is that the functor ${\rm Hom}_R(A, \_)$ is not always right exact.
Now (in the general case) we suppose that $A=Z_n$ and $H_n (\mathrm {Hom}_R(A, C)) = \{0\}.$ This means that for every $f \colon Z_n \to C_n$ such that $u_n \circ f = 0$ there exists $g \colon Z_n \to C_{n+1}$ such that $u_{n+1} \circ g = f$.
Take $f \colon Z_n \hookrightarrow C_n$ the inclusion morphism. It follows that $f(Z_n) \subset B_n$ therefore $Z_n = B_n$ i.e. $H_n = \{0\}$.