For common operators like $+$ the inverse is $-$, and $\times$ it is $\div$. Wondering if a "search" or "filter" on a set can have an inverse.
$$B = \{a \in \mathbb{Z} \mid a < 10 \land a > 5\}$$
That's a super simple search but it demonstrates the point. The inverse of a search is something like a "non-search" perhaps, but that doesn't quite make sense. A "forget" maybe. The inverse doesn't seem like search for everything "except", such as:
$$B^{-1} = \{a \in \mathbb{Z} \mid a \geq 10 \lor a \leq 5\}$$
If you are trying to "undo" a search, it seems like you just don't want to perform any search. Wondering what your thoughts are on this type of operation. The operation might look like:
$$\mathbb{Z} \circ \{a \in \mathbb{Z} \mid a < 10 \land a > 5\} = \{6, 7, 8, 9\}$$
I don't know what the inverse would look like, maybe:
$$\{a \in \mathbb{Z} \mid a < 10 \land a > 5\} \circ \{a \in \mathbb{Z} \mid a < 10 \land a > 5\}^{-1} = \mathbb{Z}$$
Or perhaps in this case, there just isn't an inverse :/. If not, wondering why certain things can't have an inverse.
Your concept of "search or "filter" is somewhat like intersection of sets. That is, if $\;A\;$ is a search space and $\;B\;$ represents a condition or criterion, then $\;C:=A\cap B\;$ represents the "search" or "filter" result. However, this clearly does not have an inverse. That is you can't recover $\;A\;$ from $\;C,\;$ the intersection even if you know $\;B\;.$ It would be a lot easier to "keep a backup" of $\;A\;$ instead.