# 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?

• Is $Z_n$ is the kernel of the map from $C_n \to C_{n-1}$ (ie the cycles)? – Tim May 31 at 19:29
• @Tim yes that is correct – BananaCats Category Theory App May 31 at 21:36
• I'm not sure I understand your question correctly, but I think you should fix a certain $n$ and understand $H_n(\operatorname{Hom}(Z_n,C))$ as the $n$th homology group of the complex $\{\operatorname{Hom}(Z_n,C_m)\}_m$. – Arnaud D. Jun 1 at 7:38

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 $$y \in Z_n$$. We would like to show that there exists an element $$x \in C_n$$ 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 $$y = i(y) = d_{n+1}(f(y))$$.