Thread: Prove inverse of a bijection is increasing

Let f: A->B be a bijection, where A and B are subsets of the set of all real numbers. Prove that if f is increasing on A, then f's inverse is increasing on B.

My professor said to use the contrapositive of "f: A->B is increasing" to prove this, but I don't understand how that would help me.

I know that if f is a bijection and g is the inverse of f, then g is also a bijection and f is the inverse of g.

I don't even know where to start.

If then .

Suppose then would happen if
.

HINT: What is

Ok I know that f(f-1(t))=t

So if f-1(s)>f-1(t) then that would mean the inverse function is decreasing on B.

Am I supposed to use f(f-1(s))<=f(f-1(t))? Can that even be true if the inverse function is decreasing?

Originally Posted by letitbemww

Ok I know that f(f-1(t))=t

So if f-1(s)>f-1(t) then that would mean the inverse function is decreasing on B. Am I supposed to use f(f-1(s))<=f(f-1(t))? Can that even be true if the inverse function is decreasing?

No it does not mean that.

It means .
What is wrong with that?

Oh, that can't be true if we already established s<=t.

Is it just that simple? I guess I was confused on how I could use the contrapositive to prove it. I don't even understand my professor's hint because there is no "if...then" statement there.

Maybe I'm just not understanding right. It seems like you are proving it by contradiction.

Almost all of this type problem can proved by contradiction.
That is what was done.

Ok thanks!!