Trying to understand proofs, help me solve this one

[IntegrateMe]

Suppose that f, g : \mathbb{R} \rightarrow \mathbb{R} are surjective (ie onto functions with domain \mathbb{R} and allowable output values \mathbb{R}). Prove that f \circ g is also surjective (ie, prove f \circ g 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?

 

[deluks917]

f being an onto function means that for every real number y there exists x such that f(x) =y.

 

[gauss^2]

f \colon A \to B means a function with name f from the domain A into the codomain B. For instance, consider the function f\colon \mathbb{R} \to \mathbb{R}\times \mathbb{R} defined by the rule f(x) = (x,x). In this example the domain (e.g., the set x is in) is the real line \mathbb{R} and the codomain is the set \mathbb{R}\times \mathbb{R}, e.g., the 2D plane. This function is not onto though, because there are points (a,b) \in \mathbb{R}\times \mathbb{R} such that there is no x \in \mathbb{R} with f(x) = (a,b). To prove this, let a, b be real with a \neq b and assume f(x) = (a,b). Since f(x) = (x,x) = (a,b) must hold, then x = a and x = b must also be true, but a \neq b so it cannot. The only parts of the codomain that are hit are those on the line y = x which can be written as L = \{(a,a)\colon a \in \mathbb{R}\}. In this case L is the range of f (it's also called the image of f), which is always a subset of the codomain.

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

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