Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
6  
$f:\mathbb R\to\mathbb R^+$, $x\mapsto|x|$, $g:\mathbb R^+\to\mathbb R$, $x\mapsto x$. –  Did Sep 18 '12 at 5:05

6 Answers 6

up vote 31 down vote accepted

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.

share|improve this answer
    
In example (1), if I am reading it correctly, $f \circ g$ is not a bijection; it is the function $\{0,1\} \to \{0,1\}$ that is always zero. The reverse composition $g \circ f$ is a bijection, but that's just the same as example (2), isn't it? –  Trevor Wilson Sep 18 '12 at 5:33
    
@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. –  Brian M. Scott Sep 18 '12 at 5:47

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.

share|improve this answer

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 $f(x)=x$.

share|improve this answer

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)

share|improve this answer

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

share|improve this answer
1  
Interestingly, your functions are bijections, but on different domains and codomains. –  akkkk Sep 18 '12 at 13:00
    
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. –  robjohn Sep 19 '12 at 15:46

X={0},Y={0,1}, f(0)=0, g(0)=g(1)=0. more info could be found at here.

share|improve this answer
6  
This is the exact same answer Brian gave 9 hours before you. –  Graphth Sep 18 '12 at 15:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.