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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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    Oh, the linked article is enough :).2011-12-17

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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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    @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.2011-12-19