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The following comes from the book Teichmüller Theory by Hubbard. During the proof of a proposition he uses the following sort of argument to say there exists a surjective homomorphism with finite kernel, but I don't understand why this proves such a map exist.

Earlier he had shown for some space $X_n^*$ that $H^1(X_n^*; \mathbb{R}) = \mathbb{R}$. He then goes on to say that since $H^1(X_n^*; \mathbb{R}) = \mathrm{Hom}(H_1(X_n^*; \mathbb{Z}), \mathbb{R})$ and since $H_1(X_n^*; \mathbb{Z})$ is a finitely-generated Abelian group, there must be a surjective homomorphism $H_1(X_n^*; \mathbb{Z}) \to \mathbb{Z}$ with finite kernel.

Why does this imply there is a surjective map with finite kernel?

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If $G$ is a finitely generated Abelian group then $G\cong\Bbb Z^m\times A$ where $A$ is a finite group and $m$ (the rank of $G$) is a nonnegative integer. Then $\textrm{Hom}(G,\Bbb R)\cong \Bbb R^m$ so for $G=H_1(X_n^*,\Bbb Z)$ the rank is $m=1$. Then there is a surjection from $G$ to $\Bbb Z$ with kernel $A$.

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