Question 1103170
Let a, b be elements of A such that 

g(f(a)) = g(f(b)).

g 1-to-1 ==> f(a) = f(b).

f 1-to-1 ==> a = b.

==> g(f(x)) is 1-to-1 also.-------------(1)

Now let z be an element of C.

g onto ==> there is y an element of B such that g(y) = z.

f onto ==> there is x an element of A such that f(x) = y. 

===> g(f(x)) = z.

===> g o f is also onto.--------------(2)

Therefore g o f is also bijective, from (1) and (2).