Inverse of Direct Image Mapping does not necessarily equal Inverse Image Mapping

Theorem
Let $S$ and $T$ be sets.

Let $\mathcal R \subseteq S \times T$ be a relation.

Let $\mathrel R^\to$ be the direct image mapping of $\mathcal R$.

Let $\mathrel R^\gets$ be the inverse image mapping of $\mathcal R$.

Then it is not necessarily the case that:
 * $\paren {\mathrel R^\to}^{-1} = \mathcal R^\gets$

where $\paren {\mathrel R^\to}^{-1}$ denote the inverse of $\mathrel R^\to$.

That is, the inverse of the direct image mapping of $\mathcal R$ does not always equal the inverse image mapping of $\mathcal R$.

Proof
Proof by Counterexample:

Let $S = T = \set {0, 1}$.

Let $\mathcal R = \set {\tuple {0, 0}, \tuple {0, 1} }$.

We have that:
 * $\mathcal R^{-1} = \set {\tuple {0, 0}, \tuple {1, 0} }$


 * $\powerset S = \powerset T = \set {\O, \set 0, \set 1, \set {0, 1} }$

Thus, by inspection:

Note that $\paren {\mathrel R^\to}^{-1}$ is the inverse of a mapping which is neither an injection nor a surjection, and so is itself not a mapping from $\powerset T$ to $\powerset S$.

This can be seen to be completely different from $\mathrel R^\gets$, which can be determined by inspection to be: