# Why are “the” morphisms of the category of topological spaces continuous maps?

In the category of topological spaces, morphisms are continuous functions and isomorphisms are called homeomorphisms.

In what sense would this make sense? Is there some abstract or technical reason why "the" morphisms "are" continuous functions, rather than something else, like functions that map open sets into open sets?

• That's the definition of this category: you define the objects and the morphisms. Your question is similar to asking why "the" field operations in R are ordinary addition and multiplication, rather than something else. Part of the way R is defined as a field is through the usual notions of adding and multiplying. It doesn't make a whole lot of sense to ask "why" we use those field operations on R, or rather you should not expect some kind of profound answer. – KCd Feb 24 '16 at 2:52
• Oh OK, so the category of topological spaces having maps that take open sets into open sets as morphisms, would be an equally valid category as the usual one. – John Smith Feb 24 '16 at 2:55
• It's also worth noting that to specify a category you need to specify both itsobjects and morphisms, but is usually (for better or worse) named after its objects only, with the morphisms implicit. The most common category of topological spaces is the one where morphisms are continuous functions, but yours (where the morphisms are now open maps) is a category too. – pjs36 Feb 24 '16 at 2:56
• @JohnSmith, yes it is an equally valid category, but it sure is not an equally useful one. – KCd Feb 24 '16 at 12:10

It's a bit like asking "Why in the set $\mathbb{N}$ the elements are $0$, $1$, $2$...?" The answer is: because we define $\mathbb{N}$ like that. If we chose different elements it would be an equally "valid" set, but it would be a different set and we wouldn't call it $\mathbb{N}$.
Some of the things this excludes are: parametric curves $[0,1] \to \mathbb{R}^2$, constant functions on any nontrivial space, the diagonal map $Y \to Y \times Y$, and embeddings $A \to A \times B$.