mapping an isomorphism b/w 2 grps

Bachelier

I googled this but couldn't find a clear answer.

Is every invertible mapping an isomorphism b/w 2 grps or does it have to be linear?

Bachelier

also does an isomorphism maps connected (separated) sets to connected (separated) sets?

jbunniii

No!
It has to be invertible AND a homomorphism, meaning it must satisfy ##\phi(ab) = \phi(a)\phi(b)##, where ##\phi## is the mapping and ##a,b## are arbitrary elements of the group. Here, the group operation is written multiplicatively. The additive version is ##\phi(a+b) = \phi(a) + \phi(b)##.

By the way, one might think that it would also be necessary to stipulate that ##\phi^{-1}## is a homomorphism, but that turns out to be automatically true if ##\phi## is a bijection and a homomorphism.

jbunniii

Are we still talking about group isomorphisms? There is no notion of "connected" or "separated" for a general group. You need to impose some additional topological structure. So what kind of groups are you working with?

Bachelier

The whole question has to deal with analysis..

if A is connected and we have T: A ---> B an isomorphism, can we say T(A) in B is connected?

I guess one still have to show that a mapping is a homomorphism even in analysis. right?

micromass

OK, but you are clearly not working with just groups. What is the structure you are working with?? What are A and B?? What kind of map is T? It's an isomorphism of what?

Bachelier

I think I was confusing the invertibilty of a Linear Mapping between 2 Vector Spaces with any function that has an inverse.

I remember in my Lin. Alg. course, we learned that if a Linear Transformation T is invertible, then it is an isomorphism between the 2 VS.

Clearly this is not the general case.