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In the case of finite-dimensional vector spaces, an endomorphism is injective if and only if it is surjective. In the case of finitely generated modules over a commutative ring, if an endomorphism is surjective, then it is injective. I am wondering about the converse, which i suspect is not true. I am interested in counterexamples that give insight into why an injective endomorphism of a finite module is not necessarily surjective.

Edit: at several occasions in abstract algebra injectivity and surjectivity appear to be dual notions. At other cases this duality breaks down. The case i am referring to seems to be one of these and i am interested in understanding what it is that makes one notion (surjectivity) "more difficult" to achieve than the other (injectivity).

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  • $\begingroup$ Well, in a category like the category of modules over a ring, I'm pretty sure it's safe to say they are dual notions all the time. The duality that breaks down is the duality of statements like "this bunch of surjections are all injections" and vice versa (but not the duality of the notions of injective and surjective.) I could be wrong, but this is how it seems to me on the surface. You're basically looking for circumstances where swapping the two notions produces different results, right? This depends on the category, not on the definition of these two types of morphisms. $\endgroup$ – rschwieb Aug 7 '13 at 16:56
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Consider the map $k[x] \to k[x]$ given by multiplication by $x$.

Edit: I feel like I can also shed some light on this duality issue. Do you know what a category is? What we are basically looking at is a category of modules. In category theory there are epimorphisms (epi's) and monomorphisms (mono's) and these are dual to each other: Every category has an opposite category and the epi's in the category correspond to the mono's in the opposite (and vice versa). So for every theorem that can be proven about epi's there is a dual version about mono's (and vice versa).

In module categories the epi's are exactly the surjections and the mono's are exactly the injections. What you've noticed is that in some categories, for instance finitely generated modules over a commutative ring, that epi implies mono. In the dual category it is true that mono implies epi, it's just that the dual category is not the module category for a commutative ring. Hence the dual statement doesn't hold for that category.

You can show that the category of finite dimensional vector spaces is its own dual. So if you prove a statement like "epi implies mono", then the dual statement "mono implies epi" is also true for that category.

So the duality you've noticed between epi's and mono's always holds. It's just it doesn't only involve a single category, it involves looking at a category and it's dual category. What you've noticed is that some categories are their own dual and some aren't.

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    $\begingroup$ I am quite familiar with category theory and i find your answer highly illuminating. Well done! $\endgroup$ – Manos Aug 7 '13 at 17:31
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    $\begingroup$ @Manos If you're interested in more duality-disappointments, try these: rings (with identity) are always projective as modules over themselves, but not always injective. All modules have an injective envelope, but they need not have a projective cover. $\endgroup$ – rschwieb Aug 8 '13 at 14:02
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For example, consider $\mathbb{Z}\to \mathbb{Z}$ multiplication by $2$.

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