# Trying to understand proofs, help me solve this one!

1. Nov 16, 2011

### 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?

2. Nov 16, 2011

### deluks917

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

3. Nov 16, 2011

### 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 $f(g(x)) = z$.

Last edited: Nov 16, 2011