Proof of a bijection to the set of subsets? For part of a proof I wanted to show that $f: \{1,2\} \to \mathcal{S}(X)$ is a bijection, where $\mathcal{S}(X)$ is the set of subsets of $X$, which in this case I know to be $\{\emptyset , X\}$. So I define $f$ as $f(1) = \emptyset$ and $f(2) = X$. But then $f(1) = \emptyset \iff 1\in f^{-1}\emptyset$, and $f$ is a bijection means $f^{-1}$ is a bijection, so $f^{-1}\emptyset = \{f^{-1}(x) : x\in \emptyset\}$. But no such x is in the empty set, so $f^{-1}\emptyset = \emptyset$, when it should have 1 as an element.
I'm confused as to what exactly is going wrong here; I'm pretty sure that $\mathcal{S}(X)$ should have size 2 so it should biject to $\{1,2\}$.
Any clarifications would be appreciated!

Be careful! $f(x)=y\iff x\in f^{-1}(\{y\})$. For every map $f\colon A\to B$ we have a map $f^{-1}\colon \mathcal P(B)\to\mathcal P(A)$, $S\mapsto\{\,x\in A\mid f(x)\in S\,\}$. If $f$ happens to be bijective, we (unfrtunately?) use the same symbol $f^{-1}$ to denote the inverse map, that maps $y\in B$ to the uniqe $x\in A$ with $f(x)=y$. You fell for this trap here (which was easily triggered because $\emptyset$ is bot an element and a subset of $\{\emptyset,X\}$.
