Answer:
Step-by-step explanation:
Given that g is a function from A to B and f is a function from B to C.
g: A -->B and f: B-->C
a) fog= f{g[x)} is a function from A to C
Let fog be onto. Then we get for any element C in f we got an image in A.
This is possible only if every element of C has an image in B because if not then f cannot be applied to g(x).
Hence proved
b) Let fog be one to one.
i.e. fog(a) = fog(b) implies a =b
This suggests that f{g(a) }=f{g(b)}
Since f is a function for g(a) and g(b) three exists unique images.
So g(a) = g(b) implies a = b.
g is one to one
c) If fog is one to one and on to then we have f is onto and g is one to one.
Since f is onto every element in g has a preimage so it follows that both f and g are bijective.