Hom$(A,C)$ is not exact?

Asked Apr 21 '21 at 15:13

Active Apr 21 '21 at 18:48

Viewed 62 times

Let $A$ be arbitrary and consider a complex $C$ of the form $$0 \to C_0 \underset{f_0}{\to} C_1 \underset{f_1}{\to} C_2 \to \cdots$$ Prove that if $C$ is exact and all the $C_i$ are free, then $\text{Hom} (A,C)$ is not exact.

I'm having trouble coming up with a counterexample for this and actually think this is a true statement. I would appreciate thoughts on proving this.

Edit: This is a chain complex where we have a sequence of homomorphisms of abelian groups shown above such that $f_{n+1} f_{n} = 0$ for each $n$. I apologize for not giving more context on this. We are discussing homology groups in Hatcher's Algebraic Topology and this is part of a study guide refresher to review complexes from Algebra. I also was told that this was a typo, the statement is in fact true but I still do not see why.

edited Apr 21 '21 at 18:48

asked Apr 21 '21 at 15:13

Dominated Convergence Theorem

2

Huh? If $A$ is allowed to be arbitrary, then I can choose $A=0$ for example. – user10354138 Apr 21 '21 at 15:18
A complex of what? – Bernard Apr 21 '21 at 15:18
@user10354138 isn't all $0$s an exact sequence? – D_S Apr 21 '21 at 15:19
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@D_S That's why I am asking for clarification – user10354138 Apr 21 '21 at 15:20
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I think OP intended to ask for an example of an $A$ for which $\operatorname{Hom}(A,C)$ is NOT exact. They should clarify – D_S Apr 21 '21 at 15:59
I edited the post to add more context and found out that the statement had a typo - the conclusion should be that we do have exactness. I apologize. – Dominated Convergence Theorem Apr 21 '21 at 18:50

Hom$(A,C)$ is not exact?

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