If f: A $\rightarrow$ B, and f is onto and one-to-one, and we have functions g: B $\rightarrow$ A and h: B $\rightarrow$ A, show g=h.

Question

The question highlights in addition that:

f $\circ$ g = f $\circ$ h = Id$_B$, and g$\circ$f = h $\circ$ f = Id$_A$

I understand that in order to prove g=h, you must show that

1. the codomain and domain are the same, and

2. g(a) = h(a) for every a $\in$ B.

This one has stumped me. I've returned to it a few times, but I can never make sense of how to go about it. Logically, it makes sense, but that obviously isn't good enough in mathematics. So any input would be great!

Answer 1

We have to show inverse of f is unique. Let, $g≠h$ Then, there exist $x\in B$ such that $g(x) ≠h(x) $.let, $g(x) =u$
and $h(x)=v$. Definitely, $u≠v$(By our assumption)

Since, $f\circ g=f\circ h$ $=Identity$ Then, $f(u)=f(v)=x$ it implies f is not one-one which contradicts the given condition. Hence, $g=h$