Example where $f\circ g$ is bijective, but neither $f$ nor $g$ is bijective Can anyone come up with an explicit example of two functions $f$ and $g$ such that: $f\circ g$ is bijective, but neither $f$ nor $g$ is bijective?
I tried the following: $$f:\mathbb{R}\rightarrow \mathbb{R^{+}}
$$ $$f(x)=x^{2}$$
and  $$g:\mathbb{R^{+}}\rightarrow \mathbb{R}$$
$$g(x)=\sqrt{x}$$
$f$ is not injective, and $g$ is not surjective, but $f\circ g$ is bijective
Any other examples?
 A: If we define $f:\mathbb{R}^2 \to \mathbb{R}$ by $f(x,y) = x$ and $g:\mathbb{R} \to \mathbb{R}^2$ by $g(x) = (x,0)$ then $f \circ g : \mathbb{R} \to \mathbb{R}$ is bijective (it is the identity) but $f$ is not injective and $g$ is not surjective.
A: If $X$ is any set at all with $\left| X \right| > 1$ then the diagonal map
$$\begin{align}\Delta : X &\to X \times X \\ x &\mapsto (x,x)\end{align}$$
and the projection map
$$\begin{align}\pi : X \times X &\to X\\ (x,y) &\mapsto x \end{align}$$
satisfy $\pi \circ \Delta = \text{id}_X$, which is bijective, and yet neither $\Delta$ nor $\pi$ is bijective.

In a similar vein, if $f : X \to Y$ is any function and $\left|Y\right| > 1$ then we can take the graph $\Gamma_f : X \to X \times Y$ given by $\Gamma_f(x) = (x,f(x))$ and the same projection. The above example is the special case where $f(x)=x$.
A: The simplest example:

*

*$X=\{0\},Y=\{0,1\}$, $f(0)=0$, $g(0)=g(1)=0$.

(Here $g\circ f$ is the bijection, since I inadvertently reversed the names of the functions.)
Everything else is an elaboration of this idea.
A: An example with the aditional restriction that there is only one set involved, i.e. we have $f\colon X\to X$ and $g\colon X\to X$ is given by 
$$f\colon\mathbb N_0\to\mathbb N_0, n\mapsto \max\{n-1,0\}$$
$$g\colon\mathbb N_0\to\mathbb N_0, n\mapsto n+1$$
(Clearly, $X$ cannot be finite, so this example is sort minimal with my additional restriction)
A: $f(x)=2x$ and $g(x)=x/2$ will do for integers (round in some way for odd numbers).
A: There hasn't been an answer with $f, g \colon \mathbb{R} \to \mathbb{R}$ yet, so consider
\begin{align*}
f(x) &= \begin{cases}
\ln x & \text{if $x > 0$}, \\
0 & \text{otherwise},
\end{cases} \\
g(x) &= e^x.
\end{align*}
(The value of $f(x)$ does not matter when $x \le 0$.)
Here, $f$ is not injective and $g$ is not surjective,
but $f \circ g = \mathrm{id}_{\mathbb{R}}$ is bijective.
A: Let $f$ and $g$ both be functions from $\mathbb N$ to $\mathbb N$ defined as follows:
$$
g(n)=2n \qquad f(m)=\begin{cases}m/2 & \text{if $m$ is even,}\\42&\text{if $m$ is odd.}\end{cases}
$$
Notice that $f\circ g$ equals the identity on $\mathbb N$. However, $g$ is not surjective – the image of $g$ equals the even numbers, and $f$ is not injective – every odd number is mapped to $42$.
