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Math Help - Show Injective/surjective

  1. #1
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    Show Injective/surjective

    f: A x b --> A defined by f(x,y) = x

    Injective: Here's what I did
    Suppose f(x,y) = f(a,b). Then x = a. Thus not injective
    Is this the correct approach?

    surjective: Not sure how to go about this... Any advice? Thanks
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  2. #2
    MHF Contributor Swlabr's Avatar
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    For injective, you need to say that b is not equal to y (you start with (a, b) \neq (x, y), but then you concluded that a=x but not that b=y).

    So, a proof would be: Note that for b \neq y, f(x, b) = f(x, y) but (x, b) \neq (x, y).

    What you did was write down the working to find the proof, but it wasn't the proof itself per se.

    For surjective, you start with something arbitrary in the image, A, and find something which maps to it. So, let x \in A. Can you think of some (a, b) \in A \times B such that f(a, b) = x?
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  3. #3
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    Quote Originally Posted by jzellt View Post
    f: A x b --> A defined by f(x,y) = x

    Injective: Here's what I did
    Suppose f(x,y) = f(a,b). Then x = a. Thus not injective
    Well, what causes the "thus"? To show "not injective", you need to assert that two different pairs, (x_1, y_1) and (x_2, y_2), can give the same value- you need to give a counterexample.

    Is this the correct approach?

    surjective: Not sure how to go about this... Any advice? Thanks
    Look at the definition of "surjective"- a function, f, from A to B, is surjective if and only if, for every y in B, there exist x in A such that f(x)= y.

    If you are given a number b does there exist a pair (x, y) such that f(x,y)= b? If so, what is it?
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