I am working through an applied category text. In it he makes several definitions:
Definition 1: Let $(P, \leq)$ be a preorder. Then an upper set for P is a set such that if $p\in P$ and $p\leq q$, then $q\in P$.
Definition 2: For a preorder as above, let $\uparrow a=\{p\in P | a\leq p \}$.
It is easy to see that the sets of definition 2 form upper sets for $P$. However, he is asking if the map $\uparrow :P^{op}\rightarrow U(P)$ is a monotone map, where $U(P)$ is the set of upper sets of $P$ with subset ordering. I tried just with the definitions, but because the orders are different I think that there may be a typo. Perhaps $U(P)$ should be $U(P^{op})$
Any help or correction would be much appreciated, thanks.
No, the order are the right way around. Let me write $\leq_{op}$ for the order in $P^{op}$, so $a \leq_{op} b$ iff $b \leq a$. We need to show that if $a \leq_{op} b$, then $\uparrow\! a \subseteq \uparrow\! b$. So let $x \in \uparrow\!a$, then $a \leq x$. Since $a \leq_{op} b$ we have $b \leq a$, and hence $b \leq a \leq x$, so $x \in \uparrow\! b$.
In words: the further we go down in $P$, the bigger the upper set becomes. So this is why we had to take the opposite order on $P$.