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Trying to understand proofs, help me solve this one!

  1. Nov 16, 2011 #1
    Suppose that [itex]f, g : \mathbb{R} \rightarrow \mathbb{R}[/itex] are surjective (ie onto functions with domain [itex]\mathbb{R}[/itex] and allowable output values [itex]\mathbb{R}[/itex]). Prove that [itex]f \circ g[/itex] is also surjective (ie, prove [itex]f \circ g[/itex] is also onto).

    First of all, I have absolutely no math theory experience, so I don't really understand what's being asked for here.

    I know that ℝ is the set of all real numbers, but I'm not sure what ℝ → ℝ represents.

    Can someone explain to me the mathematical terms and give me a breakdown of how this problem would be solved?
     
  2. jcsd
  3. Nov 16, 2011 #2
    f being an onto function means that for every real number y there exists x such that f(x) =y.
     
  4. Nov 16, 2011 #3
    [itex]f \colon A \to B[/itex] means a function with name [itex]f[/itex] from the domain [itex]A[/itex] into the codomain [itex]B[/itex]. For instance, consider the function [itex]f\colon \mathbb{R} \to \mathbb{R}\times \mathbb{R}[/itex] defined by the rule [itex]f(x) = (x,x)[/itex]. In this example the domain (e.g., the set [itex]x[/itex] is in) is the real line [itex]\mathbb{R}[/itex] and the codomain is the set [itex]\mathbb{R}\times \mathbb{R}[/itex], e.g., the 2D plane. This function is not onto though, because there are points [itex](a,b) \in \mathbb{R}\times \mathbb{R}[/itex] such that there is no [itex]x \in \mathbb{R}[/itex] with [itex]f(x) = (a,b)[/itex]. To prove this, let [itex]a, b[/itex] be real with [itex]a \neq b[/itex] and assume [itex]f(x) = (a,b)[/itex]. Since [itex]f(x) = (x,x) = (a,b)[/itex] must hold, then [itex]x = a[/itex] and [itex]x = b[/itex] must also be true, but [itex]a \neq b[/itex] so it cannot. The only parts of the codomain that are hit are those on the line [itex]y = x[/itex] which can be written as [itex]L = \{(a,a)\colon a \in \mathbb{R}\}[/itex]. In this case [itex]L[/itex] is the range of [itex]f[/itex] (it's also called the image of [itex]f[/itex]), which is always a subset of the codomain.

    However, the function [itex]g\colon \mathbb{R} \to L[/itex] defined by the rule [itex]g(x) = f(x) = (x,x)[/itex] is onto. To prove this, let [itex](a,a) \in L[/itex]. Then we see [itex]a \in \mathbb{R}[/itex] and [itex](a,a) = g(a)[/itex]. Since the element [itex](a,a)[/itex] chosen from [itex]L[/itex] was arbitrary, we see that [itex]g\colon \mathbb{R} \to L[/itex] is an onto function. In this case the range and the codomain of the function [itex]g[/itex] are the same: namely, [itex]L[/itex].

    This is really all you need to do in your problem: pick an arbitrary point [itex]z[/itex] in the codomain of [itex]f \circ g[/itex] and find a [itex]y[/itex] in the domain of [itex]f[/itex] such that [itex]f(y) = z[/itex]. Then find an [itex]x[/itex] in the domain of [itex]g[/itex] such that [itex]g(x) = y[/itex]. E.g., you find an [itex]x[/itex] such that [itex]g(x) = y[/itex] and [itex]f(y) = z[/itex] so that [itex]f(g(x)) = z[/itex].
     
    Last edited: Nov 16, 2011
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