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For an abelian group $G$ we denote by $G^*$ the $\mathbb{Z}$-module $\text{Hom}_\mathbb{Z}(G,\mathbb{Q}/\mathbb{Z})$ -- group of all $\mathbb{Z}$-module homomorphisms from $G$ to $\mathbb{Q}/\mathbb{Z}$ (the quotient group).

Now, let $A,B$ and $C$ be abelian groups. Let $0\to B\stackrel{\mu}{\to} A\stackrel{\epsilon}{\to} C\to 0$ be a sequence of homomorphisms.

Suppose $0\to C^*\stackrel{\epsilon^*}{\to} A^*\stackrel{\mu^*}{\to} B^*\to0$ is an exact sequence. Is the sequence $0\to B\stackrel{\mu}{\to} A\stackrel{\epsilon}{\to} C\to 0$ also exact?

I've heard it is, but I have no idea why. I will apreciate any hints and advices how to prove it. (I suppose that injectivity of the $\mathbb{Z}$-module $\mathbb{Q}/\mathbb{Z}$ (it is a divisible group) may be important.)

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You are looking for a condition of faithful exactness. I'll be back in a while, maybe google finds out something about it. –  tetrapharmakon Dec 17 '11 at 20:19
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Oh, the linked article is enough :). –  tetrapharmakon Dec 17 '11 at 20:35

1 Answer 1

up vote 2 down vote accepted

This statement follows from $Hom(-,I)$ being an exact functor iff $I$ is injective. To see why this is true, we will argue in the opposite category. If $I$ is injective in $\mathcal{A}$ for any category $\mathcal{A}$, then $I$ is projective in $\mathcal{A}^{op}$ by the definition of the opposite category. Now, recall that the definition of a projective module says $P$ is projective iff $Hom(P,-)$ is exact.

So we have the short exact sequence $0\rightarrow C^*\rightarrow A^*\rightarrow B^*\rightarrow 0$, which is the same as $0\rightarrow Hom(C,\mathbb{Q}/\mathbb{Z}) \rightarrow Hom(A,\mathbb{Q}/\mathbb{Z})\rightarrow Hom(B,\mathbb{Q}/\mathbb{Z}) \rightarrow 0$. Going to the opposite category, we see we get $0\leftarrow Hom(\mathbb{Q}/\mathbb{Z},C) \leftarrow Hom(\mathbb{Q}/\mathbb{Z},A)\leftarrow Hom(\mathbb{Q}/\mathbb{Z},B) \leftarrow 0$. Applying the fact that $\mathbb{Q}/\mathbb{Z}$ is projective in $\mathcal{Ab}^{op}$ and the definition of projective modules, we see that $0\rightarrow B\rightarrow A\rightarrow C$ is exact. (Note that it's important we have that $Hom(A,\mathbb{Q}/\mathbb{Z})=0 \Leftrightarrow A=0$ thus our functor $Hom(-,\mathbb{Q}/\mathbb{Z})$ is faithful.)

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I don't think this argument works. After all, $\mathbb{Q}$ is injective, too, and the image of the non-exact sequence $0 \to \mathbb{Z}/4 \to 0 \to 0 \to 0$ under $\operatorname{Hom}{(-,\mathbb{Q})}$ is simply zero, so something more than just injectivity is needed. –  t.b. Dec 17 '11 at 21:06
    
Sorry, you're right there. You need the additional property that $Hom(A,I)=0$ implies $A$ is zero for the injective object $I$. This is clear to see when we're talking about $\mathbb{Q}/\mathbb{Z}$. –  KReiser Dec 17 '11 at 21:14
    
Ah, I see. Thanks. Could you maybe edit that to your answer so that it is self-contained and correct? (you have an edit link down at the left bottom of your answer) –  t.b. Dec 17 '11 at 21:25
    
Sure, answer edited. –  KReiser Dec 18 '11 at 1:26
    
@KReiser: thank you for your answer. Unfortunately, I don't understand from it why the definition of a projective module and fact $\text{Hom}(A,\mathbb{Q}/\mathbb{Z})=0$ iff $A=0$ implies that $0\to B\to A\to C\to0$ is exact. But, I found the proof of this fact in the article linked in the comment to my question. If you have had on mind other idea than it is written there (thm 1.1, $3)\Rightarrow1)$), then please write it here. –  Damian Sobota Dec 19 '11 at 3:11

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