If $f$ is a bijection of $A$ onto $B$ show $f^{-1}$ is a bijection of $B$ onto $A$

Question

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Introduction to Real Analysis (Robert G. Bartle) 18. (b)

If $f$ is a bijection of $A$ onto $B$, show that $f^{-1}$ is a bijection of $B$ onto $A$.

Attempt:

If $f$ is a bijection $A$ onto $B$ then if $x\in A$ then $f(x)\in B$ and if $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$.

By definition of inverse, $f^{-1}:=\left\{(f(a),a): B\times A, (a,f(a))\in f \right\}$, so $f^{-1}$ is a surjection from $B$ to $A$. Moreover if $f(x_1)\neq f(x_2)$ then $x_1\neq x_2$ implying injection since $f$ is already injective hence $f^{-1}$ is injective. Since $f^{-1}$ is injective and surjective $f^{-1}$ is also bijective.

I doubt I’m correct. For one, I want the final step to be if $x_1\neq x_2$ then $f^{-1}(x_1)\neq f^{-1}(x_2)$. How do we do this?

Answer 1

For every $y\in B$ there is a unique $x_y\in A$ such that $f(x_y)=y$.

We then have $f^{-1}(\{y\})= \{t\in A : f(t) = y\} =\{ x_y\}$. So, $f^{-1}$ is a function and $f^{-1}(y)=x_y$.

If $f^{-1}(y_1) = f^{-1}(y_2)$, then ${x_{y_1}} = {x_{y_2}}$. Applying $f$ on both sides of the equality, we get $y_1 = y_2$. This proves injectivity.

Given an arbitrary $x \in A$, we have $f^{-1}({f(x)}) =x_{f(x)}=x$. This proves surjectivity.

Answer 2

Showing that if $x \neq y$ then $f^{-1}(x)\neq f^{-1}(y)$ is equivalent to prove that if $f^{-1}(x) = f^{-1}(y)$ then $x=y$. But if $f^{-1}(x) = f^{-1}(y)$ then the images by $f$ must be equal, so $f(f^{-1}(x)) = x = y = f(f^{-1}(y))$.