So suppose we have some fixed set $X$. We then consider its power set $\mathscr{P}(X)$ as a category, defined thusly:
- Objects: Members of the power set, i.e. all sets $A$ such that $A \subseteq X$.
- Arrows: Inclusion arrows.
Obviously, then, for $A, B \in \mathscr{P}(X)$, the arrow $A \rightarrow B$ exists only if $A \subseteq B$.
(Slight digression: I believe this could perhaps form some sort of poset or something with the relation basically being "this is a subset of that," since this basically implies there exists only one arrow between any two objects. Not sure if this is relevant at all, just an observation.)
My task is to prove - or disprove - that $\mathscr{P}(X) \cong \mathscr{P}(X)^{op}$.
My gut feeling is that it is isomorphic, but I'm not 100%. I think proving it would simply be a matter of making functors between the two that map each object to itself, and each arrow to its reversal in the other category, and then confirming they compose to give the identity functors ... but that seems a little too easy. So I feel like I'm overlooking some nuance; if I remember correctly it also came up in something I read a while back that categories are generally not iso to their opposites, so that also makes me wonder.
Any thoughts or nudges in the right direction?
If your set is $S$ your objects are subsets of $S$ (elements of $\mathcal{P}(S)$) and arrows are inclusions $U\subseteq V$ for $U,V \in \mathcal{P}(S)$.
To get a covariant functor from $\mathcal{P}(S)$ to the opposite category $\mathcal{P}(S)^{\text{op}}$ you can send each subset to its complement in $S$, i.e. $U \mapsto U^C = S\setminus U$.
If $U \subseteq V$ is a morphism (inclusion) in $\operatorname{Hom}_{\mathcal{P}(S)}(U,V)$ then $U^C \supseteq V^C \in \operatorname{Hom}_{\mathcal{P}(S)^{\text{op}}}(U^C,V^C)$ is the corresponding morphism in the opposite category.
This covariant functor is its own inverse, so it establishes the isomorphism.