Let $f:A\to B, g:B\to C$ and $h:C\to D$ Then show that ho(gof)=(hog)of , that is composition of functions is associative

Given $f:A\to B\text{ and }g:B\to C$ 

$\Rightarrow \text{ gof }:A\to C$ and $\text{ gof }:A\to C;h:C\to D$ 

then ho(gof) $:A\to D$ 

$g:B\to C$ and  $h:C\to D$ then hog $:B\to D$

$f:A\to B$ and $hog:B\to D$ then 

$\text{ (hog)of : }A\to D$

$\therefore$ho(gof) and (hog)of have the same domain A. 

Let $a\in A,$ then

[ho(gof )](a)$\mathit{=}h\mathit{\left[}\mathit{\right(}\mathit{\text{ gof }}\mathit{\left)}\mathit{\right(}a\mathit{\left)}\mathit{\right]}$

$=h\left[g\right(f\left(a\right)\left)\right]$

$=\left(hog\right)\left[f\right(a\left)\right]$

$=\left[\right(h o g) of ] \left(a\right)$

$\therefore$ $ho\mathit{\left(}\mathit{\text{ gof }}\mathit{\right)}\mathit{=}\mathit{\left(}hog\mathit{\right)}\mathit{\text{ of }}$