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For the exact complex $A \to B \to C$, does the complex

$$\text{Hom}(M,A) \to \text{Hom}(M,B) \to \text{Hom}(M,C)$$

fail to be exact?

If not, why do we need the exactness of $0 \to A \to B$ to get the exactness at $\text{Hom}(M,B)$?

Please teach me.

Pierre MATSUMI

Pierre MATSUMI
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  • I would suggest you try phrasing your question in a different way. As it is, I suspect it will be quickly closed. Anyway, here's a hint, what if $C = 0$? –  Jul 14 '13 at 16:15
  • This ought to be asked elsewhere, but a hint is that exactness at $A$ imposes an extra condition when it comes to exactness at $B$ (the map from $A$ onto its image in $B$ is an isomorphism). – user43208 Jul 14 '13 at 16:16

2 Answers2

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A functior $\mathcal F$ is called exact if the exactness of # $$0\to A\to B\to C\to 0$$ implies exactness of $$0\to \mathcal FA\to \mathcal FB\to \mathcal FC\to 0.$$ If we (merely) get $$0\to \mathcal FA\to \mathcal FB\to \mathcal FC,$$ the functor is called left-exact.

You assume about $\mathcal F=\operatorname{Hom}(M,\cdot)$ that exactness of $$A\to B\to C$$ always implies exactness of $$\mathcal FA\to \mathcal FB\to \mathcal FC.$$ Note that this would imply exactness of $\mathcal F$ as you can apply this to the parts $0\to A\to B$, $A\to B\to C$ and $B\to C\to 0$ and combine. The Hom-functor really is merely left-exact as can be seen from $$0\to 2\mathbb Z\to \mathbb Z\to\mathbb Z/2\mathbb Z\to 0$$ and $M=\mathbb Z/2\mathbb Z$: $$0\to0\to0\to\mathbb Z/2\mathbb Z\to 0$$ is not exact at the right end. (Actually, this is the same example as given by WillO, only extended to the full sequence).

Hagen von Eitzen
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Try $A={\mathbb Z}$, $B={\mathbb Z}/2{\mathbb Z}$, $C=0$, $M=B$.

Rasmus
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WillO
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    Those A,B,C don't from an exact sequence. – Rasmus Jul 14 '13 at 16:54
  • @Rasmus: Yes they do. – WillO Jul 14 '13 at 17:10
  • No, they don't. – Rasmus Jul 14 '13 at 18:02
  • @Rasmus: How much money you got on you? – WillO Jul 14 '13 at 18:27
  • @WillO Um, not exact at A. – Fran Burstall Jul 14 '13 at 18:31
  • @FranBurstall : The sequence is of the form $A\rightarrow B\rightarrow C$. It makes no sense to ask whether this sequence is "exact at $A$", because exactness refers to both an incoming and an outgoing arrow, and there is no incoming arrow at $A$. Exactness of this diagram means exactness at $B$, nothing more or less. – WillO Jul 14 '13 at 18:37
  • @WillO OK. Misread the OP's question. Was misled by slightly non-standard use of "exact sequence". – Fran Burstall Jul 14 '13 at 18:47
  • @FranBurstall : On the contrary, if you read the OP's question, you will see that it asks about an exact sequcne of the form $A\rightarrow B\rightarrow C$ , and then inquires whether we also need a $0$ on the left in order for $Hom$ to preserve exactness. I gave a counterexample by writing down a sequence of the form specified by the poster, which does not have a zero on the left, and where $Hom$ does not preserve exactness. That's directly responsive to the OP's question. – WillO Jul 14 '13 at 18:50
  • Just to give context to latecomers: My "On the contrary" comment addressed to Fran Burstall is a response to a comment that he subsequently deleted. I regret that Fran probably no longer wants to get in on my bet with @Rasmus, but I'm hoping Rasmus still does :) – WillO Jul 14 '13 at 18:51
  • Touche. It's exact in the middle. ;) – Rasmus Jul 14 '13 at 19:25
  • (edited to remove -1) – Rasmus Jul 14 '13 at 19:28