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.)
Best Answer
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.)