I would like someone to verify, if my below proof is correct. This may be trivial, but I would like to be sure, that I haven't overlooked anything, especially since its my first analysis course.
Let $f: X \rightarrow Y$ be bijective function, and let $f^{-1}:Y \rightarrow X$ be its inverse. Verify the cancellation laws $f^{-1}(f(x))=x$ for all $x \in X$ and $f^{-1}(f(y))=y$ for all $y \in Y$. Conclude that, $f^{-1}$ is also invertible, and has $f$ has its inverse (thus $(f^{-1})^{-1}=f$).
Proof.
(1) Claim: $f^{-1}(f(x))=x$.
$f$ is bijective. So, $f$ is both injective and surjective. Injectivity would mean, given $y \in Y$, there is at most one $x \in X$, such that $x \mapsto y$. Surjectivity means, given $y \in Y$, there is atleast one $x \in X$, such that $x \mapsto y$. Therefore, bijection means that there is a perfect matching (one-to-one correspondence) between $x \leftrightarrow y$.
Every $x$ corresponds to a unique $y$. Thus, $f(x)=y$.
Every $y$ corresponds to a unique $x$. Thus, $f^{-1}(y)=x$.
Stringing together these statements, $f^{-1}(f(x))=x$.
(2) Claim: $f(f^{-1}(y))=y$.
There is a one-to-one correspondence between $x$ and $y$ as seen above. Therefore, we infer that $f^{-1}$ is also a bijection. Using the same arguments as before, $f^{-1}(y)=x$ and $f(x)=y$. Stringing together these statements, $f(f^{-1}(y))=y$.
Best Answer
Yes. This proof checks out. Although you mentioned that $f^{-1}$ has $f$ as its inverse, I thought I'd clarify the explicit proof. By Tao's definition, ${(f^{-1})}^{-1} = f$ if for all $x\in X$, there exists a unique $y\in Y$ such that $f^{-1}(y)=x$. As you've shown, this value of $y$ is simply $f(x)$. (It's unique because $f$ is an injection)