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Let A,B be finite, commutative groups. Let $A^{*} = Hom(A, \mathbb{Q}/\mathbb{Z})$, the set of homomorphisms from $A$ to $\mathbb{Q}/\mathbb{Z}$. $A^{*}$ is abelian itself (take this for granted). Let $f: B \to A$ a homomorphism of groups. For $h \in A^{*}$ define $f^{*}h \in B^{*}$ by $f^{*}h = h \circ f$. Take for granted that $f^{*}: A^{*} \to B^{*}$ is a homomorphism.

My questions are the following. I feel I am missing something really small that will enable me to prove both really easily. Any help will be appreciated:

a) $f$ injective $\iff$ $f^{*}$ surjective

b) Switch surjective and injective above.

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1 Answer 1

up vote 1 down vote accepted

For (b) Suppose $f$ is surjective. Then you have an exact sequence

$$ A \to B \to 0$$

and applying $\textrm{Hom}(-,\Bbb{Q}/\Bbb{Z})$ gives that

$$0 \to \textrm{Hom}(B,\Bbb{Q}/\Bbb{Z}) \to \textrm{Hom}(A,\Bbb{Q}/\Bbb{Z})$$

is exact, i.e. $f_\ast$ is injective. To show the converse I think you use the fact that for a finite abelian group the dual of the dual is the group itself.

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