Constructing an inverse of a composite function

Question

Given: $A,B,C\subset \mathbb R$ and two functions $f:A\to B \;\& \;g:B\to C$ defined by $$f(x) = 2x + 1\;\;\&\;\;g(x) = \frac x3$$ Prove $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$

I've proved that g is one-one and onto.

How do I proceed further?

Answer 1

Let's first write : $$ g \circ f(x) = g(f(x)) = \frac{2x+1}{3} $$ Now we try to invert it : $$ y = \frac{2x+1}{3} \implies 3y = 2x+1 \implies \frac{3y-1}{2} = x \implies (g \circ f)^{-1}(x) = \frac{3x-1}{2} $$ Now we compute $f^{-1}$ and $g^{-1}$ : $$ 2x+1 = y \implies x = \frac{y-1}{2} \implies f^{-1}(x) = \frac{x-1}{2} \\ \frac{x}{3} = y \implies x = 3y \implies g^{-1}(x) = 3x $$ Finally we check the statement : $$ f^{-1} \circ g^{-1}(x) = \frac{3x-1}{2} = (g \circ f)^{-1}(x) $$

Answer 2

Answer by Zubzub calculates the functions and checks the identity explicitly for the given functions, which is probably what is expected. Here I give abstract reasoning why this always must hold which would be good exercise for you to try.

Hint 1: Can you define what invertible function is without mentioning bijectivity?

Answer 1:

We say that $f\colon A\to B$ is invertible if there exists $g\colon B\to A$ such that $g\circ f = \operatorname{id}_A$ and $f\circ g = \operatorname{id}_B$. Such $g$ must be unique, so we write $g = f^{-1}$.

Hint 2: Using the above definition, could you prove that for invertible functions $f\colon A\to B$ and $g\colon B\to C$ inverse of the composition is given by $(g\circ f)^{-1} = f^{-1}\circ g^{-1}$?

Answer 2:

$(g\circ f)\circ (f^{-1}\circ g^{-1}) = g\circ(f\circ f^{-1})\circ g^{-1} = g\circ \operatorname{id}_B\circ g^{-1} = g\circ g^{-1} =\operatorname{id}_C\\$ $(f^{-1}\circ g^{-1})\circ (g\circ f)= f^{-1}\circ(g^{-1}\circ g)\circ f = f^{-1}\circ \operatorname{id}_B\circ f = f^{-1}\circ f = \operatorname{id}_A\\$ $\implies(g\circ f)^{-1} = f^{-1}\circ g^{-1}$