Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Clarifying some mapping stuff

  1. Nov 5, 2012 #1


    User Avatar
    Homework Helper

    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. jcsd
  3. Nov 5, 2012 #2

    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.
  4. Nov 5, 2012 #3


    User Avatar
    Homework Helper

    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
  5. Nov 5, 2012 #4


    User Avatar
    Homework Helper

    Out of curiosity, is a bijection sufficient for an inverse or must the map also be a homomorphism for an inverse to happen.
  6. Nov 5, 2012 #5
    Being bijective is equivalent to the existence of an inverse. So yes, it is sufficient.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook