A co-topological category

Ecumenical dispatches from the London Library

A co-topological category

This post describes a nearly useless category that reminds me of the topologies associated with categories of coalgebras for a set endofunctor.

Let ${X, Y}$ be topological spaces. A not-necessarily continuous map ${f:X\rightarrow Y}$ is open if for every open set ${U\subset X}$ , ${f[U]}$ is open in ${Y}$ . The category ${\mathbf{Opt}}$ of topological spaces and open maps has the same isomorphism types as the category ${\mathbf{Top}}$ of topological spaces and continuous maps.

Proof: A map is an isomorphism in ${\mathbf{Opt}}$ if and only if it is a homeomorphism in ${\mathbf{Top}}$ . $\Box$

Coproducts exist in ${\mathbf{Opt}}$ . In general, products do not exist in ${\mathbf{Opt}}$ . However, something analogous to powers of a single space exist.

For a topological space ${X}$ , ${\square X}$ (“the square of ${X}$ “) denotes the topological space with underlying set ${X\times X}$ and topology generated by the basis

$\displaystyle \{ U\times V, \Delta_X[U] : U,V\,\mathrm{open}\,\mathrm{in}\,X\}$

where ${\Delta_X:X\rightarrow X\times X}$ is the diagonal. This topology is the smallest topology on ${X\times X}$ making the diagonal map open (and continuous).

The ${\square}$ construction defines an endofunctor ${\square}$ on ${\mathbf{Opt}}$ together with a natural transformation ${\delta_X:X\rightarrow\square X}$ (the same underlying map as ${\Delta_X}$ ) from the identity functor ${1_\mathbf{Opt}}$ to ${\square}$ . There is a one-to-one correspondence between maps ${f:X\rightarrow Y}$ and maps ${g:X\rightarrow\square Y}$ such that ${\pi_1 g= \pi_2 g}$ for ${i=1,2}$ , where ${\pi_i:\square Y\rightarrow Y}$ is the projection (projections are open).

Proof: Given a open map ${f:X \rightarrow Y}$ , the open map ${\delta_Y f}$ satisfies ${\pi_1\delta_Y f = f = \pi_2 \delta_Y f}$ . Conversely, a map ${g:X\rightarrow \square Y}$ with ${\pi_1 g= \pi_2 g}$ for ${i=1,2}$ , defines ${f=\pi_1 g}$ . Then

$\displaystyle \delta_Y f = (\pi_1 g \times \pi_1 g) \delta_Y = (\pi_1 g \times \pi_2 g) \delta_Y = (\pi_1 \times \pi_2) \delta_Y g = g$

$\Box$

Written by Christian Marks

July 6, 2011 at 3:37 AM

Posted in Bagatelle

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