| Math 323 (Laetsch) | Proving Surjective |
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Suppose we have a function f : S → T. One definition of surjective in this situation is to say that the range of f equals T. This definition, however, may lead you to believe that you have to prove something TWO ways, as you usually do when you are proving that two sets are equal (if you know how to prove that two sets are equal).Now for the point:A better way of looking at surjective is to take into account the fact that the notation f : S → T (if used correctly!) implies that T is an appropriate codomain for f, so that the range of f is already known to be a subset of T. (Of course, you have to assume that the person using the notation f : S → T is smart enough to have checked that T really is an appropriate codomain; cf. Defining Functions and Codomains.) Since we already know that the range of f is a subset of the codomain T, all we have to verify for “surjective” is that the codomain T is a subset of the range of f; i.e., every element of T is an element of the range of f. So f : S → T is surjective if and only if —
(*) For every y in T there exists x in S such that f(x) = y
(so that T is a subset of the range of f).
Consider f(x) in T.This is egregiously wrong, and it is egregiously wrong for TWO reasons --
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Last modified 3/27/08