Question

The Identity function i behaves just like the number 1. That is (F o i)=f and (i o g)= g

It is also true that for a function k, (k o k^-1)=i

use this information and algebra to solve h= (f o g) for f

any insight would be helpful!

Answer 1

I'll denote the identity function by $\mathrm{Id}$. Then for any functions $f$ with inverse $f^{-1}$,

$\begin{cases}f\circ\mathrm{Id}=f\\\mathrm{Id}\circ f=f\\f\circ f^{-1}=\mathrm{Id}\end{cases}$

One important fact is that composition is associative, meaning for functions $f,g,h$, we have

$(f\circ g)\circ h=f\circ(g\circ h)$

So given

$h=f\circ g$

we can compose the functions on either side with $g^{-1}$:

$h\circ g^{-1}=(f\circ g)\circ g^{-1}=f\circ(g\circ g^{-1})$

then apply the rules listed above:

$h\circ g^{-1}=f\circ\mathrm{Id}=f$