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`