Thread: Bijective 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.

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.

Helo !
for example if is the real numbers set and and ....

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].

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 , with , f - injective g - injective
g is not bijective g is (not) surjective

And here I stopped.

If , then Thus the restriction of to is not a bijection form H to itself.

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

Thank you.

This example may be easier:

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

Now, take . The function is not a bijection, as .

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 !

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.