Bijective functions

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March 27th 2011, 12:38 AM
veileen

Bijective functions

A is a non-empty set. Is there a bijective function $\displaystyle f:A\mapsto A$ such that there exists $H\subset A, H\neq\varnothing$ , with $\displaystyle f(H)\subset$ H, and $g:H\mapsto H$ , $g(x)=f(x)$ , $x\in H$ is not bijective?

Well, uhm, honestly, I have no idea what to do -_- Should I try to "build" a function?

Thanks in advance.
March 27th 2011, 12:47 AM
HallsofIvy

You are asked whether or not there exist a bijective function that is NOT bijective on a subset. If you believe the answer is yes, the best way to show that would be to create one. If you believe the answer is no, you will have to show why that cannot happen.
March 27th 2011, 02:03 AM
tsimaths

Helo !
for example if $A = {\mathbb R}$ is the real numbers set and $f(x)=x^3$ and $H=[\frac 12 ,1]$ ....
March 27th 2011, 03:55 AM
DrSteve

tsimaths, I don't think that example quite works. f(1/2) = 1/8 which is not in H. I think we can fix this by making H=[0,1/2].
March 27th 2011, 04:17 AM
veileen

O.O Those examples are useless. "You are asked whether or not there exist a bijective function that is NOT bijective on a subset." Better now?

We (I) suppose there is (not) a function g which is not bijective, then:

Let $x_{1}, x_{2}\in H$ , with $g(x_{1})=g(x_{2})\Leftrightarrow f(x_{1})=f(x_{2}), x_{1}, x_{2}\in H, H\subset A$ , f - injective $\Rightarrow x_{1}=x_{2} \Rightarrow$ g - injective
g is not bijective $\Rightarrow$ g is (not) surjective

And here I stopped.
March 27th 2011, 06:16 AM
DrSteve

If $f(x) = x^3$ , then $f([0,1/2])=[0,1/8].$ Thus the restriction of $f$ to $H=[0, 1/2]$ is not a bijection form H to itself.
March 27th 2011, 06:44 AM
veileen

Meow (Blush) You're right; I wasn't attentive, I'm sorry ^^'

Thank you.
March 27th 2011, 01:46 PM
topspin1617

This example may be easier:

Let $f:\mathbb{Z}\rightarrow\mathbb{Z}$ be given by $f(a)=a+1$ . This function is definitely a bijection.

Now, take $\mathbb{N}=\{0,1,2,\ldots\}$ . The function $f|_{\mathbb{N}}$ is not a bijection, as $0\notin \mathrm{im}f|_{\mathbb{N}}$ .
April 1st 2011, 12:27 PM
tsimaths

Helo

I'm sory

I want say an interval stable by $f$ such $[-a,a]$ or $[-a,0]$ or $[0,a]$ where $0 < a < 1$

Thank's DrSteve

Remark : this bijective function doesn't exist if $A$ is a finite set !
April 1st 2011, 01:14 PM
Deveno

it depends on how you define g.

if you are asking if g is a bijection on A, it needn't be.

for example, take A = [-1,1], H = [0,1] and f(x) = x, g(x) = |x|.

if you are asking if g is a bijection on H, only if f(H) = H (if f is not surjective on H, g won't be either).

example: let f(x) = x^3 A = [-1,1], H = [0,1/2], g(x) = f(x).

if you require that f be a bijection on H, then g MUST be one as well.