F maps A onto A=>f is one to one

[pacificguy]

Hi,
Say f:A->A where A is a metric space and f is onto. I think it should be true that this implies that f is also one to one. Is there a way to formally prove this?
Thanks.

 

[lurflurf]

There is no way to prove it since it is false, and it is impossible to prove untrue things.
Try to construct a counter example. Hint it is easy.

 

[Hurkyl]

Let's try starting with something simple!

The contrapositive is often something useful to consider:

f is not one-to-one ==> A is not a metric space​

What is the simplest example you can think of of an onto function that is not one-to-one? Can you prove the domain of that function cannot have a metric space structure on it?

(P.S. what is f? Any function at all?)

 

[ice109]

what does metric space even have to do with it?

i was about to construct a counter example but maybe there's something to this.

if A={a,b,c} and the range of f is all of A i don't see how one can define f in such a way that it's not injective because f(a) = a and f(a) = (b) isn't allowed right?

 

[Office_Shredder]

what does metric space even have to do with it?

i was about to construct a counter example but maybe there's something to this.

if A={a,b,c} and the range of f is all of A i don't see how one can define f in such a way that it's not injective because f(a) = a and f(a) = (b) isn't allowed right?

If A is finite and f is onto, then f must be injective

 

[ice109]

If A is finite and f is onto, then f must be injective

yea i figured that out shortly after posting