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 $d_{n+1}$?

  • $\begingroup$ 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 $\endgroup$ Jan 4, 2018 at 22:02
  • $\begingroup$ @TheMadcapLaughs in fact, the first object is fixed. $\endgroup$
    – Rafael
    Jan 5, 2018 at 2:40
  • $\begingroup$ 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 $\endgroup$
    – Jo Mo
    Aug 13, 2019 at 8:35
  • $\begingroup$ @JoBe I thought the same but immediately checked the notation in exercise 1.1.1 which uses $\mathbb Z$. $\endgroup$
    – JPhy
    Aug 5, 2020 at 16:20
  • $\begingroup$ For $A = \mathbb{Z}$ converse will hold as Hom$(\mathbb{Z}, C_n)$ is same as $C_n$ in that case. $\endgroup$
    – Kalas678
    Feb 7, 2022 at 12:57

1 Answer 1


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 $Z_n=\mathbb{Z}_2$.

  • 4
    $\begingroup$ 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). $\endgroup$ Jan 10, 2018 at 22:06
  • $\begingroup$ @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. $\endgroup$
    – Ra1le
    Jun 30, 2019 at 19:34

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