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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?

  • 7
    $f:\mathbb R\to\mathbb R^+$, $x\mapsto|x|$, $g:\mathbb R^+\to\mathbb R$, $x\mapsto x$.2012-09-18

6 Answers 6

30

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 one of this idea.

  • 0
    @Trevor: Thanks. I inadvertently had examples for both compositions, having temporarily forgotten which way round the OP had chosen, and when I got back from answering the bloomin’ phone I’d forgotten that one of them needed to be scrubbed.2012-09-18
7

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.

5

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 $f(x)=x$.

2

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)

2

$f(x)=2x$ and $g(x)=x/2$ will do for integers (round in some way for odd numbers).

  • 0
    You've mixed up $f$ and $g$: $g\circ f$ is the bijection. For maximum clarity, you should say which is not injective ($g$) and which is not surjective ($f$) or at least what the bijection is.2012-09-19
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X={0},Y={0,1}, f(0)=0, g(0)=g(1)=0. more info could be found at here.

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    This is the exact same answer Brian gave 9 hours before you.2012-09-18