Let f A rightarrow B B and g B rightarrow C be functions Pro

Let f: A rightarrow B B and g: B rightarrow C be functions. Prove: If g f is one-to-one and f is onto, then g is one-to-one. If g f is onto and g is one-to-one, then f is onto. Let A = {1, 2} and B = {a, b, c}. Let the functions f and g be f = {(1, a), (2, b)} and g = {(a, 1), (b, 2), (c, 1)}. Verify that g f = l_A, and then explain why g is not the inverse of f.

Solution

a)

Let, g(b)=g(b\')

Since f is onto so there a and a\' so that

f(a)=b,f(a\')=b\'

So, g(f(a))=g(f(a\'))

But, gof is one to one

So, a=a\'

Hence, b=b\'

HEnce, g is one to one

b)

Let, b in B

gof is onto so

There is some a in A so that

gof(a)=g(b)

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

But, g is one to one

So, f(a)=b

And b any element in B

so, f is onto

c)

gof(1)=g(f(1))=g(a)=1

gof(2)=g(f(2))=g(b)=2

Hence, gof=1_A

BEcause g is not one to one and f maps 1 only to a but g has two elements a and c mapping to 1