Just to ask a question about the alternative method for showing a group/ring homomorphism $\phi$ is bijective, hence an isomorphism, by constructing an explicit inverse $f$.

Main question: Does $f$ necessarily have to be a homomorphism as well? If yes, why?

I am a bit confused about this approach since the usual approach taught in books/ lecture so far is to show injectivity by showing the kernel is trivial, etc.

Thanks for any help.

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    $\begingroup$ No, it will automatically be a homomorphism given that it's the inverse. However, it's often possible to construct "natural" inverses that will clearly be homomorphisms by construction (e.g. because they are built out of other homomorphisms in some way). $\endgroup$ – Qiaochu Yuan Nov 2 '15 at 5:14

Let $f:A \to B$ be a bijective group homomorphism (or ring homomorphism, the argument is the same for every algebraic structure) and $g:B \to A$ be an inverse of $f$ as a function of sets. Then $$g(b_1b_2)=g(f(a_1)f(a_2))=g(f(a_1a_2))=a_1a_2=g(b_1)g(b_2)$$ where $f(a_1)=b_1$ and $f(a_2)=b_2$ exist and are unique because $f$ is bijective. Thus $g$ is necessarly a group homomorphism.

If you know a bit of category theory, you know that an isomorphism in a category is a morphism that has an inverse. What I just showed is that for algebraic categories (e.g. Groups, AbGroups, Rings, etc) it is sufficient for a morphism to be a bijection of sets. This is one of the properties that makes these algebraic categories different from the topological ones say (a continuous bijection is not necessarly a homeomorphism).

  • $\begingroup$ Thanks a lot. That will save some checking for the inverse $\endgroup$ – yoyostein Nov 2 '15 at 5:43

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