Question

(a) Let f be a one-one mapping of A onto B, g be a one-one mapping of B onto C. Then
show that gof is also invertible and
(gof)-1 = f-1 o g -1

Answer 1

Answer:

Given f:A→B is bijective

⇒f is one-one and onto.

Also, given g:B→C is bijective

⇒g is one-one and onto.

Now, we will check whether gof is bijective

Let x,y∈A such that

(gof)(x)=(gof)(y)

⇒g(f(x))=g(f(y))

⇒f(x)=f(y) (Since, g is one-one, so g(x)=g(y)⇒x=y

⇒x=y (∵f is one-one)

Hence, gof is one-one.

Now, for surjective, let z∈C be an arbitrary element

Since, g is onto, so for z∈C, there exists an element r∈B such that g(r)=z

Also since, f is one so for every x∈A, there is an element r∈B such that f(x)=r

⇒g(f(x))=g(r)=z

⇒(gof)(x)=z

Hence, for z∈C, there is an element x∈A .

Hence, gof is onto.