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If a homomorphism has a trivial kernel, is it an isomorphism?

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Not unless you assume a bunch of extra things. Try a few simple examples to see this. – Tobias Kildetoft Apr 1 '13 at 18:52
An isomorphism's image must be the whole codomain. – anon Apr 1 '13 at 18:53
I just realized that using the word "simple" might have been a bad choice, as if one interprets that as the objects being simple, that is one of the cases where the above is actually true. – Tobias Kildetoft Apr 1 '13 at 18:54
Exercise: Show that an homomorphism $f$ is injective iff has trivial kernel. But you cant conclude that $f$ is surjective. – Gastón Burrull Apr 1 '13 at 18:56
Yea your right. its only injective. thanks! – 010110111 Apr 1 '13 at 18:56
up vote 4 down vote accepted

Recall: An isomorphism is a bijective homomorphism.

We need more conditions on a homormorphism whose kernel is trivial to ensure it is an isomorphism:

A homomorphism whose kernel is trivial is injective, but not necessarily onto, hence not necessarily an isomorphism.

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More precisely, an isomorphism is a bijective homomorphism whose inverse is a homomorphism. This is not automatic in all categories, e.g., topological spaces. – lhf Dec 29 '13 at 14:46

It is not that actually. Even if you consider this identity map for example: $i:A_n \rightarrow S_n$ sending each $\tau$ to itself. $A_n$ is not isomorphic to $S_n$ right?

Ontoness is also necessary for an homomorphism to be an isomorphism. If you have an injective homomorphism which is not necessarily onto, then it is called an embedding.

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An isomorphism is a homomorphism that is injective(1-1) and surjective(onto), but the trivial kernel makes it a monomorphism not necessarily surjective.

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