# Does the box topology have a universal property?

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?

• Here's maybe a starting point: A universal property is usually defined in terms of (co)limits. Now the forgetful functor from the set has both left and right adjoints (the two simplest choices for topologies: trivial and discrete topologies for each set). Right adjoints preserve limits and left adjoints preserve colimits. Thus a universal property of the box topology would still have to be a universal property of the underlying set (and the underlying sets and functions of whatever diagram you chose to pose the universal problem that the (co)limit solves). – ffffforall Feb 1 at 5:02
• Here's a thought : the box topology actually behaves somewhat like a coproduct. Indeed, for $S$ the Sierpinski space, maps from the box to $S$ correspond to families of maps to $S$. More generally, it will be easier to find maps out of the box than into it (if there are infinitely many spaces). – Max Mar 27 at 7:43
• Here's a claim that the box topology is the product among maps "such that locally almost all members are constant". books.google.com/books?id=l-XxBwAAQBAJ&pg=PA265 – Chris Culter Mar 28 at 5:00
• @ChrisCulter: nice find! That's quite a strange claim though. It's true that families of maps $\mathbb R\to X_i$ such that locally almost all members are constant do correspond to maps from $\mathbb R$ to the box product of the $X_i.$ But obviously we can't replace $\mathbb R$ by an arbitrary space e.g. take the box product of $X_i,$ with projection maps! With the restriction to $\mathbb R$ other spaces also satisfy this property, such as the "$\aleph_1$-box topology". So it can't be a universal property. – Dap Mar 28 at 19:56