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If we have a short exact sequence of continuous group homomorphisms between abelian groups $$0 \rightarrow \mathbb{Z} \oplus \mathbb{Z} \rightarrow X \rightarrow \mathbb{Z} \rightarrow 0,$$ can we conclude that $X = \mathbb{Z}^3$? I want to say, since the map $\mathbb{Z} \oplus \mathbb{Z} \rightarrow X$ is injective by exactness, that its image is $\mathbb{Z} \oplus \mathbb{Z}$ and then just use the first isomorphism theorem; however, is that necessarily true and if so how would I be rigorous about it?

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up vote 7 down vote accepted

This as an exact sequence of abelian groups, or $\mathbb{Z}$-modules, so it is split since the third term $\mathbb{Z}$ is a projective $\mathbb{Z}$-module. By the splitting lemma, it follows $X\cong (\mathbb{Z}\oplus\mathbb{Z})\oplus\mathbb{Z}$.

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