If is bijective and is any map s.t
then is bijective.
the compostion is a function that goes from Y to Y. I am pretty sure it's onto, since f is onto. It's injective because g uniquely determines what the composition ends up as.
(we are sending to a unique function, that goes from
I think the real question I'm asking is, does this define an isomorphism?
Sorry if this is unclear. I tried my best to word it coherently. I will try to clarify if needed.
suppose g(x1) = g(x2) for x1 ≠ x2 in X.
since f is bijective, f^-1 is bijective, thus surjective.
so we can find y1,y2 in Y with f^-1(y1) = x1, f^-1(y2) = x2.
note that y1 ≠ y2 (since f^-1 exists, it is a function).
now fogof^-1(y1) = f(g(f^-1(y1))) = f(g(x1)) = f(g(x2)) = f(g(f^-1(y2))) = fogof^-1(y2),
so the composition is not injective, so it can't be bijective, either.