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I am reading Weibel's Introduction to Homological Algebra. I have come across this exercise: $$ \text{If $C$ is a chain complex and $H_n(\hom(\mathbb{Z}/n, C)) = 0$, then $H_n(C) = 0$.} $$

It seems that this is false as stated. I can take the chain complex $$ \cdots \stackrel{0}{\to} \mathbb{Z}/5 \stackrel{0}{\to} \mathbb{Z}/5 \stackrel{0}{\to} \cdots $$ Then the chain complex $\hom(\mathbb{Z}/6, C)$ has all homology groups $0$, since there are no nontrivial homomorphisms from $\mathbb{Z}/6$ to $\mathbb{Z}/5$. But, all the homology groups of $C$ are $\mathbb{Z}/5$.

J126
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1 Answers1

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He doesn’t write $\mathbb{Z}/n$. He writes $Z_n$, which is the kernel of the degree $n$ differential of the complex $C$, and nothing to do with the integers mod $n$.

Jeremy Rickard
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    Thanks. I really dislike that $\mathbb{Z}_n$ is used for $\mathbb{Z}/n$ sometimes. The exercise is now very straightforward. – J126 Oct 19 '17 at 12:58