A is a non-empty set. Is there a bijective function such that there exists , with H, and , , is not bijective?

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

Thanks in advance.

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- Mar 27th 2011, 12:38 AMveileenBijective functions
A is a non-empty set. Is there a bijective function such that there exists , with H, and , , is not bijective?

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

Thanks in advance. - Mar 27th 2011, 12:47 AMHallsofIvy
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.

- Mar 27th 2011, 02:03 AMtsimaths
Helo !

for example if is the real numbers set and and .... - Mar 27th 2011, 03:55 AMDrSteve
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].

- Mar 27th 2011, 04:17 AMveileen
O.O Those examples are useless. "You are asked whether or not there exist a bijective function that is

bijective on a subset." Better now?__NOT__

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

Let , with , f - injective g - injective

g is not bijective g is (not) surjective

And here I stopped. - Mar 27th 2011, 06:16 AMDrSteve
If , then Thus the restriction of to is not a bijection form H to itself.

- Mar 27th 2011, 06:44 AMveileen
Meow (Blush) You're right; I wasn't attentive, I'm sorry ^^'

Thank you. - Mar 27th 2011, 01:46 PMtopspin1617
This example may be easier:

Let be given by . This function is definitely a bijection.

Now, take . The function is not a bijection, as . - Apr 1st 2011, 12:27 PMtsimaths
Helo

I'm sory

I want say an interval stable by such or or where

Thank's**DrSteve**

Remark : this bijective function doesn't exist if is a finite set ! - Apr 1st 2011, 01:14 PMDeveno
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.