The proposition I would like to prove:
Proposition 1 Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be functions. Suppose $f,g$ are bijective. Then $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$
Proof. First of all, we will assume that the following lemmas are true:
Lemma 1. If arbitrary function $f: X \rightarrow Y$ is bijective, then it has an inverse.
Lemma 2. Take arbitrary bijective functions $f,g$ such that composition $g \circ f$ is defined. Then $g \circ f$ is bijective.
Now coming back to the proposition:
Since $f,g$ are bijective and $g \circ f$ is defined, by lemma 2 we know that $g \circ f$ is a bijective function. By lemma 1, we know that $(g \circ f)^{-1}$ exists.
First we show that functions $(g \circ f)^{-1}$ and $f^{-1} \circ g^{-1}$ have identical domain and codomain
Note that $g \circ f : X \rightarrow Z$ and thus $(g \circ f)^{-1} : Z \rightarrow X$. Furthermore, we have $f^{-1}: Y \rightarrow X$ and $g^{-1}: Z \rightarrow Y$, and thus $f^{-1} \circ g^{-1}: Z \rightarrow X$. We see that the $(g \circ f)^{-1}$ and $f^{-1} \circ g^{-1}$ have the same domain and the codomain.
Now we show that for arbitrary value $z \in Z$ the output of both functions will be identical.
Take arbitrary $z \in Z$. Since $g$ is surjective, we must have some $y \in Y$ such that $g(y) = z$. But since $f$ is surjective too, we must have some $x \in X$ such that $f(x) = y$.
So we have $g(f(x)) = (g \circ f)(x) = z$ and therefore $(g \circ f)^{-1}(z) = x$. Now since $g(y) = z$, $g^{-1}(z) = y$. And also $f^{-1}(y) = x$. Implying that $f^{-1}(g^{-1}(z)) = x = (f^{-1} \circ g^{-1})(z)$.
Hence $f^{-1} \circ g^{-1} = (g \circ f)^{-1}$. $\Box$
Is it correct?
Best Answer
Your proof is correct.
Alternatively you can do the following: Note that if the inverse of a function exists, then it's unique. Actually this should be included in Lemma 1, and the proof is similar to what you did in the last paragraph.
Now combining this fact and Lemma 2, and noting that $$(g\circ f)\circ (f^{-1}\circ g^{-1})=g\circ (f\circ f^{-1})\circ g^{-1}=\mathrm{id}_Z,$$ and $$(f^{-1}\circ g^{-1})\circ(g\circ f)=f^{-1}\circ(g^{-1}\circ g)\circ f=\mathrm{id}_X,$$ we can conclude that $(g\circ f )^{-1}=f^{-1}\circ g^{-1}$.