Does the box topology have a universal property?

Question

Given a set of topological spaces $\{X_\alpha\}$, there are two main topologies we can give to the Cartesian product $\Pi_\alpha X_\alpha$: the product topology and the box topology. The product topology has the following universal property: given a topological space $Y$ and a family $\{f_\alpha\}$ of continuous maps from $Y$ to each $X_\alpha$, there exists a continuous map from $Y$ to $\Pi_\alpha X_\alpha$. Now the box topology does not have this universal property, but my question is, does it have some other universal property?

On a related note, does there exist some category whose objects are topological spaces and whose morphisms are something other than continuous maps, such that the Cartesian product endowed with the box topology is the correct product object in that category?

Answer 1

This is an answer only in a special case.

An Alexandrov space is a topological space such that open subsets are closed under arbitrary intersections. They constitute a full subcategory $\mathsf{Alex}$ of $\mathsf{Top}$.

The box topology provides products in the category $\mathsf{Alex}$.

This can be either checked directly, or using the isomorphism $\mathsf{Alex} \cong \mathsf{Pre}$ with the category of preorders and the component-wise description of products of preorders.

Answer 2

I'll denote the box product topology by $\prod^\square$.

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Box topology as a limit in $\mathbf{Top}$

If we wish to remain in the category of topological spaces and continuous maps, things aren't so pretty in general, but here's a (possibly cheating?) categorical characterisation of the box product. Let $\Sigma = \{0,1\}$ be the set of truth values and endow it with the the Sierpiński topology. For an indexing set $I$, consider the set $\Sigma^I$ of functions $I\to\Sigma$ and endow it with the initial topology induced by the conjuncion map $\wedge:\Sigma^I\to\Sigma$ (explicitly, the only nontrivial open set of $\Sigma^I$ is the singleton containing the map $\operatorname{const}_1:\Sigma^I\to\Sigma$).

Now, let $(X_i)_{i\in I}$ be a family of topological spaces, and consider the following diagram:

• there is a vertex $X_i$ for every $i\in I$
• for every family of continuous functions $(f_i:X_i\to\Sigma)_{i\in I}$, add a copy of $\Sigma$ and $\Sigma^I$ connected by an edge $\Sigma^I\xrightarrow\wedge\Sigma$, and draw edges $X_i\xrightarrow{f_i}\Sigma$ for all $i\in I$

The limit of this diagram will be $X := \prod_{i\in I}^\square X_i$. Indeed, a cone for this diagram consists of

• a projection map $\pi_i:X\to X_i$ for all $i\in I$
• for every family $(f_i:X_i\to\Sigma)_{i\in I}$, a map $f:X\to\Sigma^I$ such that $\require{AMScd}$ \begin{CD} X @>f>> \Sigma^I \\ @V\pi_iVV @VV\wedge V \\ X_i @>>f_i> \Sigma \end{CD} commutes for all $i\in I$. (This map $f:X\to\Sigma^I$ sends a tuple $(x_i)_i$ to the function mapping $i\mapsto f_i(x_i)$.)

Notice that a family of open maps $f_i:X_i\to\Sigma$ is a choice of open sets $U_i\subseteq X_i$, and the open set corresponding to $\wedge f:X\to\Sigma^I\to\Sigma$ is $\prod_iU_i$, so this is really just an elaborate way of specifying the base of open sets generating $X$. To see that $X$ really is univerally a cone for the above diagram, suppose $Z$ is another cone. From the projections $g_i:Z\to X_i$, we have a necessarily unique map $g:Z\to X$ of sets factoring through the projections $\pi_i:X\to X_i$ thanks to the underlying set of $X$ being the product of the underlying sets of the $X_i$'s, so we just have to check that $g$ is continuous.

It is enough to check the preimage of basis opens in $X$, so let $U_i\subseteq X_i$ be open for all $i$, then this corresponds to a family of maps $f_i:X_i\to\Sigma$. By the fact that $Z$ is a cone, we have a corresponding map $\tilde f:Z\to\Sigma^I$ such that $\wedge\tilde f=f_ig_i$ for all $i\in I$. Since the preimage $g^{-1}\prod_iU_i$ corresponds to the map $Z\xrightarrow gX\xrightarrow f\Sigma^I\xrightarrow\wedge\Sigma$, and this map is equal to the composite $Z\xrightarrow{\tilde f}\Sigma^I\xrightarrow\wedge\Sigma$ of continuous functions, we can conclude that $g^{-1}\prod_iU_i$ is open in $Z$, as desired.

Remark. The shape of the diagram has exactly one $X_i$ for each $i$ as a leg, so the limit of the diagram can be thought of as some sort of restriction of the Cartesian product (the restrictions being defined by all the outgoing arrows of the diagram) similar to how a pullback $A\times_CB$ is a restriction of the product $A\times B$ whose restrictions are dictated by the maps $A\to C\gets B$. This is reflected also by the induced (identity-on-elements) map $\prod_i^\square X_i\to\prod_iX_i$.

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Box topology as a Cartesian product

Consider the category $\mathcal T$ where

• the objects are topological spaces
• the morphisms $X\to Y$ are (not necessarily continuous) functions $f:X\to Y$ such that for any $U\subseteq X$ open, there exists open $V\subseteq Y$ such that $f(U)\subseteq V$ and $f(X)\setminus f(U)\subseteq Y\setminus V$.

There is an issue with this definition, as I don't think composition makes sense in $\mathcal T$.

Fix a family $(X_i)_{i\in I}$ of topological spaces, then their product is given by $X := \prod_{i\in I}^\square X_i$. Indeed, the projections $\pi_i:X\to X_i$ are open, and are therefore maps in $\mathcal T$, so suppose we have morphisms $f_i:Z\to X_i$ for all $i\in I$. We have a necessarily unique function $f:Z\to X$ compatible with the projections as before, so we need only check that this is an arrow in $\mathcal T$.

Let $U\subseteq Z$ be open, then for every $i\in I$ we have some $V_i\subseteq X_i$ open such that $f_i(U)\subseteq V_i$ and $f_i(Z)\setminus f_i(U)\subseteq X_i\setminus V_i$. Set $V := \prod_{i\in I}V_i\subseteq X$.

• $V$ is open in $X$ by the definition of the box topology
• if $z\in U$, then $f_i(z)\in V_i$ for every $i$ and thus $f(z)=(f_i(z))_{i\in I}\in\prod_{i\in I}V_i=V$
• if $x\in f(Z)\setminus f(U)$, then $x=f(z)$ for some $z\in Z$ with $f(z)\notin f(U)$. Therefore, $f_i(z)\notin f_i(U)$ and thus $f_i(z)\notin V_i$ for some $i$, meaning that $x=f(z)\in X\setminus V$

Therefore, $f$ is an arrow of $\mathcal T$, as desired.

Remark. An arrow $f:X\to Y$ in $\mathcal T$ is an isomorphism iff it is a homeomorphism. Indeed, $f$ being an isomorphism in $\mathcal T$ makes it bijective, and therefore given an open $U\subseteq X$, we have $f(X)\setminus f(U)=Y\setminus f(U)$, which forces that our choice of $V\subseteq Y$ to be $V = f(U)$. Since $V$ has to be open, this means that $f$ is an open map. On the other hand, since $f$ is an isomorphism, its inverse $g:B\to A$ must also be open. This implies that for $V\subseteq Y$ open, $g(V)=f^{-1}(V)$ must also be open. As this makes $f$ a continuous open bijection, it is a homeomorphism.

In particular, since the box topology is defined as a product in $\mathcal T$, it is unique up to unique isomorphism in $\mathcal T$, and is thus unique up to unique homeomorphism, meaning that this is really a statement about the topology of $\prod_{i\in I}^\square X_i$.