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A problem I'm attempting says

Let $p:A \to B$ be a map of sets and $p^*: \mathcal{P}B \to \mathcal{P}A$ be the induced map of power sets sending $X \subseteq B$ to $p^*(X) = \{a \in A: p(a) \in X\}$. Exhibit left and right adjoints to $p^*$

but I can't quite work out what it's saying the functor is: namely, are we

$(i)$ defining $p^*$ to be the functor's action on any given map $p$ between any two sets, i.e. $(-)^*$ our functor takes a function $p$ to a function $p^*$ (and $p$ is an object, not a morphism - perhaps morphisms would be commutative squares of functions), or

$(ii)$ defining $p^*$ as a functor which acts as the power set operator on sets $(A \mapsto \mathcal{P}(A))$ for any $A$, and acts on any function between sets by $p \mapsto p^*$ (so I suppose this would be a functor from Set$^\text{op}$ to Set), or

$(iii)$ fixing sets $A$ and $B$ in advance, and just defining $p^*$ on all functions $A \mapsto B$, rather than all functions between any two sets (I am not sure that this actually works, but perhaps you can formalise a single set as a single-object category with functions to itself as morphisms, a little like you can with a single group), or

$(iv)$ none of the above?

So in essence I suppose I am asking how we actually consider $p^*$ as a functor. Are our sets fixed in advance? I don't need you to actually do the question for me, I'm sure I can do that myself once I understand what's being asked, but I would appreciate your thoughts on why the question actually means what you say it does, e.g. why the other options wouldn't work (or wouldn't be sensible).

I suspect it's meant to be $(ii)$ since $p^*(X) = p^{-1}(X)$ so this would probably behave nicely with regards to functoriality, but some confirmation would be good, since I often come across these slightly ambiguously worded category theory problems so being able to quickly interpret what the question actually means would be very useful.

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up vote 4 down vote accepted

$A$, $B$ and $p:A \to B$ are fixed. Then the powersets $\mathcal{P}A$ and $\mathcal{P}B$ are partially ordered by inclusion, and as all posets can be regarded as categories (in $\mathcal{P}A$ objects are subsets of $A$ and between two subsets $X$ and $Y$ there is exactly one morphism if $X \subseteq Y$ and no morphisms otherwise). And $p^*$ is a functor from the category $\mathcal{P}B$ to $\mathcal{P}A$, the formula you give defines it on objects (i.e. subsets of $B$), and its action on morphisms is forced by the fact that both domain and codomain are categories that come from partially ordered sets. Namely, where does $p^*$ send the unique morphism from $X_1$ to $X_2$? Well, it would have to be to the unique morphism from $p^*(X_1)$ to $p^*(X_2)$ if there is such a morphism (and there is because $X_1 \subseteq X_2$ implies $p^*(X_1) \subseteq p^*(X_2)$).

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That's great, thanks. Was that obvious from the phrasing of the question? (It wasn't to me!) – Spyam May 16 '12 at 15:36
It does seem to be consistent with the original question. – M Turgeon May 16 '12 at 15:38
Actually in hindsight I think I was probably overthinking things - once you know to add the poset structure it does become fairly clear. – Spyam May 16 '12 at 15:40
@Spyam We will see what Omar answers, but one reason why this could be "obvious" from the phrasing, is that $p^*$ is presented as a functor, whereas in (ii), $p^*$ would be the image of $p$ under your functor. Moreover, it seems $A$, $B$, and $p$ are fixed, and not allowed to vary. – M Turgeon May 16 '12 at 15:41
M Turgeon is right: I think this is the right interpretation because the question asks for adjoints to $p^*$, implying that $p^*$ is a functor itself, and not the value of some other functor on the morphism $p$. – Omar Antolín-Camarena May 16 '12 at 15:55

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