1. Feb 14, 2008

### Piffo

If I want to prove that a non specified function f(x) that maps x -->x' is onto could I show that f(x) is one to one and that f(x')^-1 (the inverse function) is also one to one??
Would that be a valid justification to say that thus f(x) must be onto?

More specifically I am looking to prove that every strictly increasing function is onto.

Francesco

2. Feb 14, 2008

### Manchot

Not every strictly increasing function is onto (e.g., f(x)=x+u(x), where u is the Heaviside step function).

3. Feb 14, 2008

### HallsofIvy

Proving that a function is one-to-one tells you nothing about it being "onto". In particular, you can't use the inverse function because if you don't already know that a function is "onto", you don't know that it has an inverse.

You can stop "looking to prove that every strictly increasing function is onto" because it is not true. The function for R to R defined by y= x if x< 0, y= x+1 if $x\ge 0$ is strictly increasing but is not "onto".

4. Feb 14, 2008

### Manchot

Wow. We came up with the same counterexample. :-)

5. Feb 14, 2008

Perhaps you are trying to prove that every strictly increasing function has an inverse mapping its range back to it's domain or something like that. Or, in other words, that if x<>y, then f(x)<>f(y).

6. Feb 14, 2008

### EnumaElish

There is the possibility that the OP meant "bijective" by "1-to-1."
http://en.wikipedia.org/wiki/Bijection