1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Abstract algebra proof: composition of mappings

  1. Sep 29, 2005 #1
    define right-inverse of a mapping B to be mapping A, such that B * A= identity (iota). Where the operation * is composition. Note that B is A's left-inverse.


    Assume S is a nonempty set and that A is an element of M(S) -the set of all mappings S->S.

    a) Prove A has a left inverse relative to * iff A is one-one
    b) Prove that A has a right inverse relative to * iff A is onto.

    I answered a) to the best of my ability, using firstly a theorem that states (B*A is 1-1) -> A is 1-1. Then, I simply constructed B from A (since A is 1-1) to prove the converse.

    b) on the other hand, i found a little harder. Once again, i used a theorem that said A*B is onto -> A is onto. Now, though, I can't seem to prove the converse. The question is, how can i construct B, knowing only that A is onto? any help would be greatly appreciated.
  2. jcsd
  3. Sep 29, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    B will not be unique. For any s in S, define A-1({s}) as {s' in S : A(s') = s}. Pick a mapping B that satisfies [itex]B(s) \in A^{-1}(\{s\})[/itex]. Then A*B(s) = A(x) for some x in A-1({s}) = {s' in S : A(s') = s}. So x is some element of S whose image under A is s, so A(x) = s, as desired. Some examples:

    1) A is also 1-1. Well in this case, |A-1({s})| = 1 for every s in S, so B is unique. You would denote the unique element of A-1({s}) simply as A-1(s).

    2) S = R, A(x) = xsin(x). Then A-1({0}) = {2k[itex]\pi[/itex] : k in Z}. You have many choices for B(0), i.e. the B(0) = 2k[itex]\pi[/itex] is a valid choice for each different k in Z. And you'll have many available choices for B(x) for each x in R. In this case, you can choose B(x) to be the smallest non-negative element of A-1({x}). You could check that this is a well-defined choice (that A-1({x}) does have a smallest non-negative element, unlike the set {1, 1/2, 1/3, 1/4, 1/5, ...}) but that would probably not be directly related to what you're studying.

    In general, however, A might be some very messy, unnatural function, and there may be no obvious or generalizable choice for B(x), but it's hard for me to give you an example because for A to be onto but not 1-1, S must be infinite, and for me to define a function A on an infinite set is only reasonably possible when A is not messy. If it followed no obvious rule, then I'd just have to list all the pairs (x, A(x)) and I couldn't possibly do that.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook