Making sense of a map, given a new domain and codomain

Question

I'm having trouble making sense of a definition from Seven Sketches in Compositionality.

Exercise 1.66. Let (P, ≤) be a preorder, and recall the notion of opposite preorder from Example 1.58.

1. Show that the set $\uparrow p := \{p' \in P | p ≤ p'\}$ is an upper set, for any p ∈ P.
2. Show that this construction defines a monotone map $\uparrow: P_{op} \rightarrow U(P)$

...

In (1), they're defining a map $\uparrow: P \rightarrow \mathcal{P}(P)$. Or perhaps it's more accurate to say they're defining a family of maps (or a way to construct maps).

Now, if we are asked to use this construction to "define a map" on a new domain (as (2) seems to be doing), there seems to be only one obvious way to do so: $\uparrow: Q \rightarrow \mathcal{P}(Q) := \{q' \in Q | q ≤_{Q} q'\}$. Applying this to the domain $P_{op}$, we get:

$$\uparrow : P_{op} \rightarrow \mathcal{P}(P_{op})$$ $$p \mapsto \{p' \in P_{op} | p \leq_{op} p'\}$$

But then the monotonicity condition is false. For example, $3 \leq_{op} 2$ but $\{3, 2, 1\} \nsubseteq \{2, 1\}$.

What they apparently want is for us to use $\leq$ when constructing the upper set, but $\leq_{op}$ when checking monotonicity. Somehow this possibility never even crossed my mind. Even knowing now that this is the correct interpretation, I literally cannot read it that way.

Should it have been obvious? How would I know to interpret the question this way, other than the fact that it's false otherwise (which I wouldn't be able to detect before interpreting the question)? Is there another way to phrase their question to make it more precise and unambiguous?

I'm having tremendous difficulty getting through the text because of descriptions like this (see also here, and also my confusion here from another book).

(Edited to be more painfully clear what I'm asking.)

Answer 1

In (1) we are indeed taking about an arrow $\uparrow: P \to \mathcal{P}(P)$. Assuming that $U(P)$ denotes the set of upper sets on $P$, we have that $U(P) \subseteq \mathcal{P}(P)$, and the image of $\uparrow$ is included in $U(P)$. So we may view it as a map $\uparrow: P \to U(P)$.

Now, $P$ is a pre-order by assumption. The powerset $\mathcal{P}(P)$ carries a natural order, namely the inclusion (making it into a partial order). Then $U(P)$ inherits this order. There is a link between the order on $P$ and the one on $U(P)$. For any $p, q \in P$ we have $$ p \leq q \implies \uparrow q \subseteq \uparrow p. $$ This is precisely the wrong way around for $\uparrow$ to be monotone. So we just flip the order on $P$ to make $\uparrow$ monotone: $$ q \leq_{op} p \implies \uparrow q \subseteq \uparrow p. $$ Note that we did not change the definition of the map $\uparrow$, so we still calculate $\uparrow p$ with respect to the original order.

If you know any category theory, then you know that we can view $P$ and $U(P)$ as categories. This is then saying precisely that $\uparrow: P \to U(P)$ is a contravariant functor.