Consider f A B and g B A suppose f g is identity functions

Consider f : A B and g : B A. suppose f g is identity functions (on B). show that:

a)f is a surjection

b)g is an injection

Solution

f: A to B and g: B to A

Hence fog: A to A

If fog is identity function

then fog (x) = x or f = g inverse and g = f inverse

i.e. Both f and g have inverse.

fog(x) is f{g(x)} is unique only if g is one to one.

Hence g is an injection.

If for some y in g g inverse cannot be found in f means

fog(x) cannot be identity.

Hence f has to be onto