[" Let "f:A rarr B" and "g:B rarr C" be the "],[" bijective functions.Then (gof) ' is "]

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

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

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

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

Let f:A rarr B and g:B rarr C be two functions.Then; if gof is onto then g is onto; if gof is one one then f is one-one and if gof is onto and g is one one then f is onto and if gof is one one and f is onto then g is one one.

Let f : A rarr B and g : B rarr C be onto functions, show that gof is an onto function.

Let f:A rarr B and g:B rarr C be two functions and gof:A rarr C is define statement(s) is true?

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.