# Exercise 1.1.4, Weibel

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 $$d_{n+1}$$?

• Notice that $id_{Z_n}(z)$ is an element of $C_{n}$, not of $ker(d_n^*)$ (the latter is a set of arrows, not of elements). The hypothesis $H_n(Hom(Z_n,C_n))=0$ actually tells you that every arrow $f:Z_n\rightarrow C_n$ such that $d_n f=0$ can be seen as a composition $d_{n+1}g$ for some other arrow $g:Z_n\rightarrow C_{n+1}$: now you only need to take the right $f$ and you're done Jan 4, 2018 at 22:02
• @TheMadcapLaughs in fact, the first object is fixed. Jan 5, 2018 at 2:40
• Oh wow, I've been trying to solve this exercise for over 30 minutes now, and I couldn't because I didn't think and hence thought $Z_n = \mathbb{Z}/(n\mathbb{Z})$ lmao Aug 13, 2019 at 8:35
• @JoBe I thought the same but immediately checked the notation in exercise 1.1.1 which uses $\mathbb Z$.
– JPhy
Aug 5, 2020 at 16:20
• For $A = \mathbb{Z}$ converse will hold as Hom$(\mathbb{Z}, C_n)$ is same as $C_n$ in that case. Feb 7, 2022 at 12:57

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 $Z_n=\mathbb{Z}_2$.
• Thank you for the counterexample! So it is not true because I would get a new complex: $0\rightarrow 0\rightarrow 0\rightarrow \mathbb{Z}_2\rightarrow 0$, and $\frac{\operatorname{ker}(\mathbb{Z_2}\rightarrow 0)}{\operatorname{im}(0\rightarrow\mathbb{Z}_2)}\cong\mathbb{Z}_2\neq 0$? I was also looking for when the converse holds (I've updated the question). Jan 10, 2018 at 22:06
• @cansomeonehelpmeout, $H_n(C_\bullet) = 0 \ \Rightarrow \ H_n(\operatorname{Hom}(Z_n,C_\bullet)) = 0$ implies, that any endomorphism $f : Z_n \to Z_n$ factorizes through $C_{n+1}$, i.e. there is a map $g : Z_n \to C_{n+1}$ such that $d_{n+1} \circ g =f$. In particular, $d_{n+1}$ has a right inverse (this is a necessry condition). Having $H_n(C_\bullet) = 0$ you can always construct a (set) function $g : Z_n \to C_{n+1}$ such that $d_{n+1} \circ g =f$. Simply let $g(x) = y$, where $d_{n+1} (y) = f(x)$. Problems start when you want to show that $g$ is an $R$-module. Jun 30, 2019 at 19:34