Let `f:A to B and g:B to C` be bijection, then `(fog)^(-1)`=

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

Let f:A to A and g:A to A be two functions such that fog(x)=gof (x)=x for all x in A Statement-1: {x in A: f(x)=g(x)}={x in A: f(x)=x}={x in A: g(x)=x} Statement-2: f:A to A is bijection.

Let f:A to A and g:A to A be two functions such that fog(x)=gof (x)=x for all x in A Statement-1: {x in A: f(x)=g(x)}={x in A: f(x)=x}={x in A: g(x)=x} Statement-2: f:A to A is bijection.

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

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.

Show that if f: A to B and g : B to C are one-one then fog: A to C is also one-one.

If f: A to B, g:B to C are two bijective functions then prove that gof:A to C is also a bijective function.

Let f: A to B; g: B to A be two functions such that fog = I_B . Then; f is a surjection and g is an injection.