Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathfrak{A}$ is a poset. For $a, b \in \mathfrak{A}$ we will denote $a \curlyvee b$ if only if there is a non-least element $c$ such that $c \leqslant a \wedge c \leqslant b$.

Let $\mathfrak{A}$ and $\mathfrak{B}$ are posets. A pointfree funcoid from $\mathfrak{A}$ to $\mathfrak{B}$ is a pair $\left( \alpha ; \beta \right)$ of functions $\alpha \in \mathfrak{B}^{\mathfrak{A}}$ and $\beta \in \mathfrak{A}^{\mathfrak{B}}$ such that $\alpha \left( x \right) \curlyvee y \Leftrightarrow \beta \left( y \right) \curlyvee x$ for every $x \in \mathfrak{A}$, $y \in \mathfrak{B}$.

I denote $\left\langle \left( \alpha ; \beta \right) \right\rangle = \alpha$ for every pointfree funcoid $\left( \alpha ; \beta \right)$.

Composition $\left( \alpha_1 ; \beta_1 \right) \circ \left( \alpha_0 ; \beta_0 \right)$ of funcoids $\left( \alpha_0 ; \beta_0 \right)$ and $\left( \alpha_1 ; \beta_1 \right)$ is defined by the formula: $$ \left( \alpha_1 ; \beta_1 \right) \circ \left( \alpha_0 ; \beta_0 \right) = \left( \alpha_1 \circ \alpha_0 ; \beta_0 \circ \beta_1 \right) . $$ The category of pointfree funcoids is the category whose objects are small posets, whose morphisms are pointfree funcoids between these posets, the composition is the composition of pointfree funcoids.

It is easy to verify that it is indeed a category.

Question: If $f$ is an isomorphism of the category of pointfree funcoids then $\left\langle f \right\rangle$ is an order isomorphism?

You can read more about funcoids (a tool for general topology) here.

share|cite|improve this question

Obviously $\left\langle f \right\rangle$ is increasing.

We have: $\left\langle f \right\rangle \circ \left\langle f^{- 1} \right\rangle = \left\langle f \circ f^{- 1} \right\rangle = \left\langle \operatorname{id}^{\mathsf{\operatorname{FCD}}}_{\mathfrak{B}} \right\rangle =\operatorname{id}_{\mathfrak{B}}$ and $\left\langle f^{- 1} \right\rangle \circ \left\langle f \right\rangle = \left\langle f^{- 1} \circ f \right\rangle = \left\langle \operatorname{id}^{\mathsf{\operatorname{FCD}}}_{\mathfrak{A}} \right\rangle = \operatorname{id}_{\mathfrak{A}}$. Thus $\left\langle f \right\rangle$ is a bijection.

$\left\langle f \right\rangle$ is increasing and bijective. Consequently $\langle f\rangle$ is an order isomorphism.

share|cite|improve this answer
Err, the condition that $\langle f \rangle$ is increasing seems not always true for every pointfree funcoid. We should additionally require this condition. I will check this. – porton Jul 21 '12 at 14:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.