# Functions real analysis

1. Sep 19, 2011

### Punkyc7

Let f and g be functions such that (g$\circ$f)(x)=x for all x $\epsilon$D(f) and (f$\circ$g)(y)=y for all y $\epsilon$D(g). Prove that a g = f^-1

Pf/

How would you go about starting this besides saying

Let f and g be functions such that (g$\circ$f)(x)=x for all x $\epsilon$D(f) and (f$\circ$g)(y)=y for all y $\epsilon$D(g).

isn't obvious that the functions would have to be the inverse of each other, How else could you get the identity? So how do you prove it, can I just say clearly it is.

Last edited: Sep 19, 2011
2. Sep 19, 2011

### benorin

What does g = f^-1 mean? My calc text gives this definition:

g = f^-1 if, and only if, f(x) is a one-to-one function with domain D(f) and range D(g) such that

g(y)=x <=> f(x)=y for all y in D(g).

3. Sep 19, 2011

### Punkyc7

g= f^-1 means that g is the inverse of f

4. Sep 19, 2011

### Punkyc7

How do you prove a definition true. If its a definition shouldn't it always be true?

5. Sep 20, 2011

### Punkyc7

This is what I have been able to get to

(g$\circ$f)(x)=x for all x ϵD(f) and (f$\circ$g)(y)=y for all y ϵD(g).

So choose
g(y)=x and f(x)=y

so

x=g(y)=f^-1(y)

ans since f and g are bijections we can conclude g=f^-1

does that work