Weibel exercise 1.1.4, taking $A = Z_n$... 
Exercise 1.1.4 Show that $\{\text{Hom}_R(A, C_n)\}$ forms a chain complex of abelian groups for every $R$-module $A$ and every $R$-module chain complex $C_{\cdot}$.  Taking $A = Z_n$, show that if $H_n(\text{Hom}_R(Z_n, C)) = 0$, then $H_n(C) = 0$. 
  Is the converse true?
Weibel, Charles A.. An Introduction to Homological Algebra (Cambridge
  Studies in Advanced Mathematics) . Cambridge University Press. Kindle
  Edition.

Do they mean take $A = Z_n$ for some single, fixed $n$?  How could they mean otherwise since the functoriality of the $\text{Hom}_R(A, \cdot)$ is only guaranteed when you hold the other argument constant.  Yet they go on to talk about $H_n$ as if that safer assumption is not true.
So which is it, or is there a mathematical mistake here?
 A: As @Arnaud D mentioned, we should think of $n$ as fixed. We are applying $Hom_R(Z_n,\square)$ to the complex $C$ and then looking at the $n$th (same $n$) homology of the result.
Let the complex $C$ be as follows:
$$\cdots\xrightarrow{} C_{n+1} \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1}\xrightarrow{} \cdots$$
Apply $Hom_R(Z_n,\square)$ and assume the $n$th homology is zero. That is, for the maps
$$Hom_R(Z_n,C_{n+1}) \xrightarrow{d_{n+1}^*} Hom_R(Z_n,C_n) \xrightarrow{d_n^*} Hom_R(Z_n,C_{n-1})$$
we have that $\ker d_n^*$ = Im $d_{n+1}^*$.
Now, let $y \in Z_n \subseteq C_n$. We would like to show that there exists an element $x \in C_{n+1}$ such that $d_{n+1}(x)=y$. Notice that the inclusion map $i: Z_n \to C_n$ is an element of the middle term above. Since $Z_n = \ker d_n$, we have that $d_n \circ i = 0$, that is, $i \in \ker d_{n}^*$. Since $\ker d_n^* =$ Im $d_{n+1}^*$ by assumption, there exists a map $f: Z_n \to C_{n+1}$ such that $d_{n+1}\circ f = i$. But then we are done since $y = i(y) = d_{n+1}(f(y))$.
