Given 2 functions $f,g:\mathbb{R} \to \mathbb{R}$ with $g \circ f$ surjective. Show that $f$ does not need to be surjective.

Question

A friend has given me this task and asked for help. I immediately tried picking various $f$ and $g$ such that $g$ and $g\circ f$ are both surjective but $f$ is not. However, i always found myself restricting $\mathbb{R}$ to some subsets $A \subset \mathbb{R}$ in order to make it work. I can't figure out how to prove it for

$$ \mathbb{R} \xrightarrow{f} \mathbb{R} \xrightarrow{g} \mathbb{R}$$

Is it even possible? I also tried using sectionwise defined functions (different functions for $x \ge 0$ and $x < 0$). However, no success.

Any advise? I'm starting to assume that it's not possible for $\mathbb{R}$ without any restrictions.

Thanks for any help!

Answer 1

Take, for instance,$$f(x)=\begin{cases}x&\text{ if }x\leqslant-1\\x+3&\text{ otherwise}\end{cases}\text{ and }g(x)=x^3-3x.$$It is clear that $f$ is not surjective. However,$$g\bigl(f(\mathbb{R})\bigr)=g\bigl((-\infty,-1]\cup(2,+\infty)\bigr)=\mathbb R.$$

Answer 2

For all real numbers $x$ we have $\log(e^{x})=x$. Take $f(x)=e^{x}, g(x)=\log\, x$ if $x>0$ and $g(x)=0$ for $x\leq 0$.

Answer 3

Consider $g:(0,1) \rightarrow \mathbb{R}$

$g(x)= \begin{array}{cc} \{ & \begin{array}{cc} \frac{1}{x} - 2 & x < \frac{1}{2} \\ \ \frac{-1}{1-x} + 2 & x >= \frac{1}{2} \\ \end{array} \} \end{array} $

and define $f: \mathbb{R} \rightarrow (0,1)$

$f(x)=\frac{1}{1+e^x}$ Is there a bijective map from $(0,1)$ to $\mathbb{R}$?

$g o f$ is surjective but $f$ is not.