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

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

Let f : A to B and g : B to C be the bijective functions. Then (g of )^(-1) is

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

Let f: A rarr B and g : B rarr C be the bijective funtions. Then (gof)^-1 is

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 : A to B and g: B to C are two bijective functions then prove that gof : A to C is also a bijection.

If f : A to B and g : B to C are two injective functions the prove that gof : A to C is also an injection.

If f : A rarr B is a bijective function, then f^(-1) of is equal to

If f : A rarr B is a bijective function, then f^(-1) of is equal to