If f and g are two bijections; then gof is a bijection and `(gof)^-1 = f^-1 o g^-1`

If f and g are two bijections; then gof is a bijection and (gof)^(-1)=f^(-1)og^(-1)

If f: A-A ,g: A-A are two bijections, then prove that fog is an injection (ii) fog is a surjection.

If f:X to Y and g:Y to Z be two bijective functions, then prove that (gof)^(-1) =f^(-1)og(-1) .

If f:AtoB and g:BtoC are both bijective functions then (gof)^(-1) is

If A={1,2,3},B=(alpha,beta,gamma),c=(p,q,r) and (f:A to B,g:B to C are defined by f={(1,alpha),(2,gamma),(3,beta)},g={(alpha,q),(gamma,p)} then show that f and g are bijective functions and (gof)^(-1)=f^(-1)og^(-1).

Let f : X rarr Y and g : Y rarr Z be two invertible functions. Then gof is also invertible with (gof)^–1 = f^ –1ofg^–1

If f: A-> B and g: B-> C be the bijective function, then (gof)^(-1) is: