Let $f:A\to B,I_A$ and $I_B$ be identity functions on A and B respectively. Then show that $fo{I}_A=f=I_B\mathit{\text{of }}$

To prove $fo{I}_A=f:$ 

We have $I_A:A\to A$ and $f:A\to B$ are functions.

$\therefore fo{I}_A:A\to B$is a function.

$\therefore fo{I}_A$ and  $f$ have same domain A 

Let $a\in A$ then $\left(fo{I}_A\right)\left(a\right)=f\left(I_A\left(a\right)\right)$ 

$=f\left(a\right) \left(\because I_A\left(a\right)=a\forall a\in A\right)$

$\therefore f_O{I}_A=f\dots ..\left(1\right)$

To prove $I_{B}of=f:$

$f:A\to B$ and $I_B:B\to B$ are two functions

$\therefore I_B\text{ of }:A\to B$ is a function

$\therefore I_B\text{ of }$ and $f$ have same domain A. 

Let $a\in A$ then $\left(I_B\text{ of }\right)\left(a\right)=I_B\left(f\right(a\left)\right)$

$=f\left(a\right)\left[\because I_B\text{ beidentity }\right] \therefore I_B\text{ of }=f\dots \dots .\left(2\right)$

$\text{ From (1) \& (2) }$

$fo{I}_A=I_B\text{ o}f=f$