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Homework Help: Graph of a function only if first projection is bijective

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    If A and B are sets, prove that a subset [itex]\Gamma\subset A X B[/itex] is the graph of some function from A to B if and only if the first projection [itex]\rho: \Gamma\rightarrow A[/itex] is a bijection.

    2. Relevant equations

    3. The attempt at a solution

    I first thought that i should define bijection by saying that a a bijection exists when there is both injection and surjection.

    If [itex]\rho: \Gamma\rightarrow A[/itex] is not a bijection then it is either
    1)not surjective
    2)not injective
    3)both 1) and 2)

    I thought that i should prove that [itex]\Gamma[/itex] is not the graph of some function A -> B when the first projection is not bijective by showing the non-surjective and non-injective cases separately. And then prove that it is, in fact, [itex]\Gamma[/itex] is the graph of the function when the first projection is bijective.

    So, i have to show the cases for when the first projection is not injective, when it's not surjective, and then when it is bijective.
    But, how do i prove that the [itex]\Gamma[/itex] is either the graph of the function or not? in any of the 3 cases?

    I just don't know where to start. Is this the right approach or is there a shorter way to do this?
  2. jcsd
  3. Oct 21, 2011 #2
    You're on the right way. Answer these:

    What does it mean for [itex]\Gamma[/itex] to be the graph of a function. What could go wrong?? (there are two things that could go wrong). What is the projection in the case that it goes wrong??

    Maybe try some examples of [itex]\Gamma[/itex] first to get used to it.
  4. Oct 23, 2011 #3
    what does a graph of a function mean. Well, doesn't it just mean that it represents the relationship between the domain and the co domain? and what could go wrong..?

    Well, i feel like the only things that can go wrong are if there are two values of codomain for one value of domain. Or, there just might be values in the co domain that is not the image of any value in the domain.

    I mean, don't these two things just mean, the function is not surjective and injective?
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