Given a poset $(P,\leq)$ the interval topology $\tau_i(P)$ on $P$ is generated by $$\{P\setminus{\downarrow x} : x\in P\} \cup \{P\setminus{\uparrow x} : x\in P\},$$ where $\downarrow x = \{y\in P: y\leq x\}$ and $\uparrow x = \{y\in P: y\geq x\}$.

It turns out that every order-preserving function is continuous with respect to the interval topology. So we get a functor ${\bf F}$ from ${\bf Poset}$, the category of partially ordered sets with order-preserving maps, to ${\bf Top}$, the category of topological spaces with continuous functions.

Does ${\bf F}$ have a left adjoint? Does it have a right adjoint?


I think the premise of the question is mistaken because $\mathbf{F}(f)$ need not be continuous for a poset map $f: P \to Q$, i.e., $\mathbf{F}$ is not functorial. I was about to ask about this in a comment, but trying to flesh it out in a comment is too space-consuming.

Consider the case where $f = \pi_1: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is projection onto the first coordinate. I claim that any finite intersection of subbasis elements for the interval topology on $\mathbb{R} \times \mathbb{R}$ cannot fit inside a half-plane $\pi_1^{-1}((-\infty, 0))$, so this half-plane is not open in the interval topology.

Subbasis elements in the interval topology for $\mathbb{R} \times \mathbb{R}$ are "L-shaped" (possibly rotated), i.e., complements of type

$$U_{a, b} = \neg\{(x, y): (x, y) \leq (a, b)\}$$ or

$$V_{a, b} = \neg\{(x, y): (a, b) \leq (x, y)\},$$ and so they correspond to conditions $(x > a) \vee (y > b)$ (for $U_{a, b}$) and $(x < a) \vee (y < b)$ (for $V_{a, b}$). Consider a finite intersection of such subbasis elements, say

$$U_{a_1, b_1} \cap \ldots \cap U_{a_m, b_m} \cap V_{a_{m+1}, b_{m+1}} \cap \ldots \cap V_{a_{m+n}, b_{m+n}}. \qquad (*)$$

If there are no $V$'s, only $U$'s, then the intersection $(*)$ contains an L-shape $U_{\max(a_i), \max(b_i)}$ which, being an L-shape, is too big to fit inside a half-plane. Same if there are no $U$'s, only $V$'s.

Suppose now that we have a mix of $U$'s and $V$'s. By distributing conjunctions over disjunctions, such a finite intersection as displayed above may be re-expressed as a union of $2^{m+n}$ sets, each corresponding to a conjunction of $m+n$ basic strict inequalities (each a condition of the form $x < a$ or $x > a$ or $y < b$ or $y > b$). One such conjunction is

$$(x > a_1) \wedge \ldots \wedge (x > a_m) \wedge (y < b_{m+1}) \wedge \ldots \wedge (y < b_{m+n})$$

which is equivalent to the condition $(x > \max\{a_1, \ldots, a_m\}) \wedge (y < \min\{b_{m+1}, \ldots, b_{m+n}\})$, which defines a quarter-plane in a southeast quadrant. Similarly, another such conjunction is equivalent to to the condition $(x < \min\{a_{m+1}, \ldots, a_{m+n}\}) \wedge (y > \max\{b_1, \ldots, b_m\})$ which defines a northwest quadrant. The intersection $(*)$ thus contains a union of a southeast quadrant and a northwest quadrant, and this union is too big to fit entirely inside the half-plane $\pi_1^{-1}((-\infty, 0)$ (which corresponds to the condition $x < 0$).

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    $\begingroup$ Well, I posted this seconds after MTyson posted a comment citing an answer of the OP. That was a lot of wasted effort on my part! $\endgroup$ – Todd Trimble Sep 24 '17 at 17:03
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    $\begingroup$ If you admit the axiom 'useful to at least one reader '$\Rightarrow$'$\neg$(a lot of wasted effort)', then this answer was not 'a lot of wasted effort'. $\endgroup$ – Peter Heinig Sep 24 '17 at 18:14

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