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Show that if U and T are onto, then UT is also onto

  1. Feb 14, 2017 #1
    1. The problem statement, all variables and given/known data
    Same as title, where T maps from a set V to W, and U maps from W to Z.

    2. Relevant equations


    3. The attempt at a solution
    I know that the standard definition for a function being onto is that for all elements w in the codomain, there exists an element v in the domain such that T(v) = w. I am not sure exactly how to concretely apply this to show that U composed with T is also onto.

    To start, we have U(T(v)) = z, and we need to show that for all z in Z there exists a v in V such that that mapping is true. First we can use the fact that U is onto, which means there is always a w = T(v) that maps to z, and since T is onto, there is always a v that maps to w. Is this sufficient to show that UT is also onto? Is there a way to write this argument using less words?
     
  2. jcsd
  3. Feb 14, 2017 #2

    fresh_42

    Staff: Mentor

    You confused the order a bit.

    To start let's see what we have (which is always a good idea): ##\;V \stackrel{T}{\twoheadrightarrow} W \stackrel{U}{\twoheadrightarrow} Z##.
    A map ##f: X \twoheadrightarrow Y## is surjective, if the following is true (you have already said this, but you asked for a short way to write it):
    $$ \forall \; y \in Y \, : \, f^{-1}(y) = \{x \in X\,\vert \,f(x)=y\} \neq \emptyset \stackrel{or}{\Longleftrightarrow} \forall \; y \in Y \; \exists \; x\in X \, : \, f(x)=y$$
    Now we have to start with an arbitrary point ##z \in Z##. The ##v\in V## for which ##U(T(v))=z## holds, has to be shown to exist.

    So first, you have to get back from ##Z## to ##W## by ##U\; -\;## no ##v##, no ##V## and no ##T## at this stage, because all you have is an arbitrary element ##z \in Z## for which you need to show ##(U \circ T)^{-1}(z) \neq \emptyset \; .##
    If you have your element in ##W##, then you can proceed using ##T##, but not earlier.
     
  4. Feb 14, 2017 #3
    But U is surjective, so there must be a w in W s.t. U(w) = z, for any z in Z
     
  5. Feb 14, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, and for those ##w## you can repeat the argument now with ##T##.
     
  6. Feb 14, 2017 #5
    So, basically, we need to show that for all z in Z there exists a v in V s.t. UT(v) = U(T(v)) = z, and we do this by first noting that there must be a w in W s.t. U(w) = z, and also that there must be a v in V s.t. T(v) = w, and hence there must be a v in V s.t. UT(v) = z? For my purposes is this an okay way of putting the argument?
     
  7. Feb 14, 2017 #6

    fresh_42

    Staff: Mentor

    Yes. I would drop the word surjective in both cases to justify the existence of those elements, but yes, that's it.
     
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