Composition of functions implies equality

[netcaster]

We have three functions: f:A->A, g:A->A and h:A->A
with both f and g bijective and h bijective.

We know that f ° h = h ° g for every x in A.

Is it true that f=g for every x in A?

I have tried to solve it and I am pretty sure it is true but I cannot find neither a counterexample nor a simple proof.

 

[HallsofIvy]

It took me just a few minutes to come up with a counter-example:

Let A= {1, 2, 3}, f= {(1,3), (2, 1), (3, 2)}, g= {(1, 2) , (2, 3), (3,1)}, h= {(1, 3), (2, 2), (3, 1)}.

 

[netcaster]

Many thanks! Your solution is simple and hence great!

 

[netcaster]

Your example is based on permutations..

What if h is only surjective rather than bijective?

The counterexample is not anymore valid...