Usually they're not bijective. Suppose, for example, that $X_1,X_2,X_3,\ldots$ is an infinite sequence of random variables, each with the same distribution, and $\Pr(X_1=0)=\Pr(X_1=1)=1/2$. Now consider just the first three. We have $ (X_1,X_2,X_3)=\begin{cases} (0,0,0) & \text{with probability }1/8 \\ (0,0,1) & \text{with probability }1/8 \\ (0,1,0) & \text{with probability }1/8 \\ (1,0,0) & \text{with probability }1/8 \\ (0,1,1) & \text{with probability }1/8 \\ (1,0,1) & \text{with probability }1/8 \\ (1,1,0) & \text{with probability }1/8 \\ (1,1,1) & \text{with probability }1/8 \end{cases} $ Here you have eight disjoint subsets of $\Omega$, and in four of them, one has $X_1(\omega)=0$. So $X_1$ is certainly not one-to-one! Nowhere near it. And if one considers the first four random variables in this sequence, one further splits each of these eight subsets in half, getting 16 subsets, and so on. So there are infinitely many $\omega\in\Omega$ such that $X_1(\omega)=0$. Very far from bijective!
The set $X_1^{-1}(\{0\})$ is precisely $\{\omega\in\Omega : X_1(\omega)=0\}$. It's an infinite set.
Generally, $f^{-1}(A)$ is defined as $\{x : f(x)\in A\}$. So the notation $f^{-1}$ does not refer to an inverse function in this context.
Sometimes in set theory one uses a notation with square brackets, writing $f^{-1}[A]$, and similarly $f[A]= \{f(x): x\in\text{domain}\}$. The reason for that is that set theorists often deal with sets that are simultaneously a member and a subset of the domain, so that $f(A)$ differs from $f[A]$.