# Clarifying some mapping stuff

1. Nov 5, 2012

### Zondrina

So I want to clarify if what I'm thinking is correct.

Suppose we have a mapping f : A → B and we have a in A and b in B.

If f is an injective map, then f(a) = f(b) implies that a = b or conversely a≠b implies f(a)≠f(b).

If f is a surjective map, then for b in B, there exists an a in A such that f(a) = b.

If A = B then f is a homomorphism from A to B if it is operation preserving. That is f(ab) = f(a)f(b) for all a and b in A.

If f is both injective, surjective, and operation preserving, then it is a bijective homomorphism, also known as an isomorphism, and thus has an inverse f-1 : B → A.

If f is an injective homomorphism, it is called a monomorphism.

If f is a surjective homomorphism, it is called an epimorphism.

If A = B and f is a homomorphism, then it is called and endomorphism.

Also a bijective endomorphism is an automorphism.

I'm hoping that those are correct ^

2. Nov 5, 2012

### micromass

Correct.

It depends. What are A and B?? Are they groups? rings? modules?? You should say that. If A and B are groups, then your definition is correct. But we usually call that a group homomorphism (although we use homomorphism when the structure of group is understood).
Also, I see no reason why we should take A=B.

That is all correct.

3. Nov 5, 2012

### Zondrina

Yes I intended for A and B to be groups. The only reason I took A = B is to imply that f was mapping from a group to itself.

Last edited: Nov 5, 2012
4. Nov 5, 2012

### Zondrina

Out of curiosity, is a bijection sufficient for an inverse or must the map also be a homomorphism for an inverse to happen.

5. Nov 5, 2012

### micromass

Being bijective is equivalent to the existence of an inverse. So yes, it is sufficient.