Are injective structure preserving maps between structures that are known to be isomorphic always surjective as well?

In my particular case, I’m curious about injective module homomorphims between two modules that I know are isomorphic, but I’m wondering whether (if true) the statement can be generalized? I haven’t been able to prove it for modules, and it seems (at least intuitively) that if a counterexample exists it would be strange. Thanks in advance!

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This fails for modules in general. For example a direct sum of an infinite number of copies of some module. You will need suitable finiteness conditions for this to work.

– Tobias Kildetoft

Oct 20 at 18:18

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1 Answer

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No, there’s no reason. The typical issue is an issue of size: for infinite-dimension vector spaces / modules, an injective endomorphism is not necessarily an isomorphism. For example if your base ring is RRR, take the direct product RNR^\mathbb{N} of N\mathbb{N} copies of RR, then the morphism RN→RNR^\mathbb{N} \to R^\mathbb{N} given by

(x0,x1,…)↦(0,x0,x1,…)(x_0, x_1, \dots) \mapsto (0, x_0, x_1, \dots)

is clearly injective but not surjective.

You can basically reuse this example in tons of category (the map N→N\mathbb{N} \to \mathbb{N}, n↦n+1n \mapsto n+1 is injective but not surjective, etc, etc). I’m not really aware of a name for a category that would satisfy the property you mention.

This is exactly what I was looking for, thanks! If we were to restrict the condition to finitely generated modules instead would the statement hold?

– stochasm

Oct 20 at 18:29

Not even then, now that I think of it… Consider Z→Z\mathbb Z \to \mathbb Z, n↦2nn \mapsto 2n, this is injective but not surjective. However over a PID the converse is true (finitely generated + surjective implies injective).

– Najib Idrissi

Oct 20 at 20:14