TITLE: Exercise 1.1.4, Weibel
QUESTION [8 upvotes]: I'm new to homological algebra, and I'm trying to learn it using Weibel's An introduction to homological algebra. I'm currently having trouble with exercise 1.1.4:

Show that $\{\operatorname{Hom}_R(A, C_n)\}$ 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 $H_n(\operatorname{Hom}_R(Z_n,C))=0$, then $H_n(C)=0$. Is the converse true?

I think the first part is OK, using that if the chain complex looks like:
$$C:~\ldots\rightarrow C_n\overset{d_n}{\rightarrow}C_{n-1}\rightarrow\ldots$$
Then we get a new chain complex:
$$C^*:~\ldots\rightarrow\operatorname{Hom}(A,C_n)\overset{d^*_n}{\rightarrow}\operatorname{Hom}(A,C_{n-1})\rightarrow\ldots$$
Where $d^*(f_n)=d_n\circ f_n$ for a $f_n\in\operatorname{Hom}(A,C_n)$. This is a chain complex since $d^*_{n-1}\circ d^*_n(f_n)=(d_{n-1}\circ d_n)(f_n)=0$.
I'm not so sure about the second part. Assuming $Z_n=\operatorname{ker}(d_n)$, and $\frac{\operatorname{ker}(d^*_n)}{\operatorname{im}(d^*_{n+1})}=0$. 
I need to show that $\operatorname{ker}(d_n)\subseteq \operatorname{im}(d_{n+1})$. Let $z\in\ker(d_n)$, then $\operatorname{id_{Z_n}}\in\operatorname{Hom}(Z_n,C_n)$ since $Z_n\subseteq C_n$. $\operatorname{id_{Z_n}}\in\ker(d^*_n)=\operatorname{im}(d^*_{n+1})$, $z=\operatorname{id_{Z_n}}(z)=d_{n+1}\circ f_{n+1}(z)$ for some $f_{n+1}\in \operatorname{Hom}(Z_n,C_n)$, so $$\ker(d_n)\subseteq\operatorname{im}(d_{n+1})$$ So $H_n(C)=0$. Is this reasoning correct?
I'm also having trouble with the converse:

$\operatorname{im}(d^*_{n+1})\subseteq\ker(d^*_n)$ is OK
Let $f_n\in\ker(d^*_n)$, then $d_n\circ f_n(z)=0$ and so $f_n(z)\in\ker(d_n)=\operatorname{im}(d_{n+1})$. Therefore $f_n(z)=d_{n+1}(y)$ for some $y\in C_{n+1}$. I think I'm done if I can find a morphism $$g:Z_n\rightarrow C_{n+1}$$ such that $g(z)=y$, but I'm having trouble arguing when this is possible (if it is). 

Any help is much appreciated!
(Edit: $\operatorname{id}_{Z_n}\in\operatorname{ker}(d^*_n)$ is an arrow, and not an element)

Are there any criteria for when the diagram above commutes? For instance, by hypothesis $Z_n\cong B_n=\operatorname{im}(d_{n+1})$, so do I need $g$ to be surjective which would mean $g$ is the inverse of $d_{n+1}$?

REPLY [8 votes]: Elements of $\ker(d_n^*)$ are arrows. Let $i_n:Z_n\rightarrow C_n$ be the canonical inclusion. Then $i_n\in Hom(Z_n,C_n)$ and
$$d^*_n(i_n)=d_n\circ i_n=0.$$
Hence there exists $u\in Hom(Z_n,C_{n+1})$ such that $i_n=d_{n+1}\circ u$ and therefore
$$Z_n\subseteq B_n=im(d_{n+1})$$
Note that this argument is valid to any exact category.
The reciproque is false: consider the complex $0\rightarrow 2\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}_2\rightarrow0$ and take $Z_n=\mathbb{Z}_2$.
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