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Is a surjective homomorphism between two (abstractly) isomorphic finitely generated abelian groups necessarily an isomorphism? I know this is true if the groups are torsion (finite) or torsion-free. What about for rings?

The question I really care about is the following. I have a deg 1 map $f: X \rightarrow Y$, where $X,Y$ are oriented manifolds, so that $f_*: H_*(X; \mathbb{Z}) \rightarrow H_*(Y; \mathbb{Z})$ is a surjective map. I also know that $H_*(X; \mathbb{Z}) \cong H_*(Y; \mathbb{Z})$. So I am wondering whether this implies that $f_*$ is an isomorphism.

user39598
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  • Hint: use the classification of finitely generated abelian groups. Try taking a quotient and seeing whether it's possible to end up with what you started with... – Dylan Wilson Jan 14 '16 at 12:16

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This is true for any finitely generated module $M$ over a commutative noetherian ring $A$: $M$ is then itself a noetherian $A$-module. If $f\colon M\rightarrow M$ is a surjective $A$-linear map, then the chain of $A$-submodules of $M$ $$ \ker(f)\subseteq \ker(f^2)\subseteq\dotsb\subseteq \ker(f^n)\subseteq\dotsb $$ becomes stationary as $M$ is noetherian. Hence there is some $n$ such that $\ker(f^n) = \ker(f^{n+1})$. Now, take $x\in \ker(f)$. Since $f$ is surjective, $f^n\colon M\rightarrow M$ is also surjective, so there exists $y\in M$ such that $x = f^n(y)$. This implies $$ 0 = f(x) = f^{n+1}(y) = f^n(y) =x, $$ where we have used that $\ker(f^{n+1}) = \ker(f^n)$. This shows $\ker(f)=\{0\}$.

Hence every surjective $A$-linear map $f\colon M\rightarrow M$ of a noetherian $A$-module $M$ is an isomorphism.

Claudius
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