Suppose function $f$ is a composition of invertible functions. Is $f$ always invertible? Is $f^{-1} $ the composition in reverse order?

Question

By an " invertible function $f$ " I mean " a function such that its inverse relation $f^{-1}$is also a function" .

Example : Let $f $ be a function such that $f(x)= \frac {x+3} {x-2}$.

We have : $f(x)= \frac {x+3} {x-2} )= \frac { x-2 + 5} {x-2}= 1 + \frac {5} {x-2}$.

So $f(x) = h(i(j(k(x)))) $

with

$k(x)= x-2$

$j(x)= x^{-1}$

$i(x)=5x$

and

$h(x)= x+1$.

It happens that $f^{-1} (x)$

$=\frac{2x+3}{x-1}$

$= (\frac {5} {x-1}) +2$

$= (\frac{x-1}{5})^{-1} +2$

$ = k^{-1}(j^{-1}(i^{-1}(h^{-1}(x))))$.

Is this true as a general rule?

Answer 1

Yes this is right, just use induction on number of functions (let's call it $n$).

• ($n=1$) This is trivial, if $f$ is invertible, therefore $f$ is invertible. :)
• Let $g(x)$ be $f_2(f_3(...f_n(x)...)$. We know $g(x)$ is invertible by induction hypothesis. So we need to show that and $h=f_1(g(x))$ is invertible. But we know $f_1(x)$ is invertible. so we can construct the inverse easily. It's just $h^{-1}=g^{-1}(f_1^{-1}(x))$. Here's the proof:

$h^{-1}(h(x)) = g^{-1}(f_1^{-1}(f_1(g(x)))) = g^{-1}(g(x)) = x$

$h(h^{-1}(x)) = f_1(g(g^{-1}(f_1^{-1}(x)))) = f_1(f_1^{-1}(x)) = x.$