# Inverse function proof

1. Mar 30, 2009

### kathrynag

1. The problem statement, all variables and given/known data

If f:x-->y has an inverse function g, then g:y--->x is one to one and onto

2. Relevant equations

3. The attempt at a solution
Let g be the inverse of f:x--->y
I think this must have something to do with an isomorphism because of one to one and onto.

2. Mar 30, 2009

### waht

has more to do with the function being bijective which is both one-one and onto (injective and surjective)

just have to work the definitions, one-one means that only one distinct element gets mapped, and onto implies that the entire range has to be mapped, together these conditions guarantee that only one element gets uniquely mapped to another element throughout the domain we are working in.

3. Mar 30, 2009

### kathrynag

Ok, I understand those defintions. My problem is understanding why g has to be bijective.

4. Mar 30, 2009

### e(ho0n3

Do you know how to prove that a function is bijective?

5. Mar 30, 2009

### kathrynag

show a(x)=a(y) implies x=y
onto, not entirely sure. I know it has something to do with showing entire range is used

6. Mar 30, 2009

### e(ho0n3

Not the range, but the codomain, i.e. for every y in the codomain, there is an x in the domain with y = f(x).

7. Mar 30, 2009

### kathrynag

Let g be the inverse of f:y--->x

Then g(a)=g(b)
a=b
Does this prove one to one? For some resaon I think it doesn't.

8. Mar 30, 2009

### e(ho0n3

There is an intermediate step between "g(a) = g(b)" and "a = b". How would get rid of the g in g(a) to get a?

9. Mar 30, 2009

### kathrynag

I'm blanking on that

10. Mar 30, 2009

### HallsofIvy

Staff Emeritus
Apply the inverse of g to both sides of the equation.

11. Mar 30, 2009

### kathrynag

so g^(-1)g(a)=g^(-1)g(b)?

12. Mar 30, 2009

### e(ho0n3

If g is the inverse of f, then what is the inverse of g? Also, what does g^{-1}(g(a)) evaluate to?

13. Mar 30, 2009

### kathrynag

f is the inverse of g?

14. Mar 30, 2009

### e(ho0n3

15. Mar 30, 2009

### kathrynag

f(g(a))

16. Mar 30, 2009

### e(ho0n3

And that simplifies to ...

17. Mar 30, 2009

### kathrynag

a I think

18. Mar 31, 2009

### e(ho0n3

That's right. Now what is the complete proof that g is one-to-one?

19. Mar 31, 2009

### kathrynag

by showing a=b