Discreteness of topological category?

Question

In his definition 18.38 Dan Freed requires the maps of a topological category to satisfy the algebraic relation of a discrete category. Is that restriction to discrete category just for convenience or the "discreteness" will play an important role somewhere?

Answer 1

Freed mentions that these relations are described in §13 of his notes, just after 13.8. The point of this reference is to concisely say that a topological category is like an ordinary category, but with topological spaces instead of sets. I don't believe the notion of discreteness is crucial afterwards.

Calling these maps “algebraic relations” highlights that, just as a group or a ring can be defined as a set and some functions obeying some constraints, a (small) category can be defined as two sets, some functions, and some constraints:

• a set $C_0$ of objects, and a set $C_1$ of morphisms.
• a map $i:C_0\to C_1$ sending every object $X$ to the identity morphism $\mathrm{id}_X:X\to X$.
• a source map $s: C_1\to C_0$ sending a morphism $f:X\to Y$ to $X$, and similarly a target map $t:C_1\to C_0$ sending $f$ to $Y$, and
• a composition map $c:C_1\times_{C_0} C_1\to C_1$. The fiber product means that we can compose when the target of the first map agrees with the source of the second; then, this map sends $(f,g)$ to $g\circ f$.

The algebraic relations force composition to be associative, $\mathrm{id}_X$ to be the identity for composition, etc. Specifically,

• $s\circ i = t\circ i = \mathrm{id}_{C_0}$: the identity map $\mathrm{id}_X$ must go from $X$ to $X$.
• $c(f,c(g,h)) = c(c(f,g), h)$, so composition is associative,
• $s(c(f,g)) = s(f)$ and $t(c(f,g)) = t(g)$, so the domain of $g\circ f$ is that of $f$, and the codomain is that of $g$, and
• $c(f,i(X)) = c(i(X), f) = f$, so composing with the identity does nothing.

In the same way, one might define a group to be a set $G$ with an associative binary operation $f:G\times G\to G$, an identity $e*\to G$, and an inverse (set map): $i:G\to G$. Then, to define a topological group, one replaces the set $G$ with a space $G$ and the set maps $f$, $e$, and $i$ with continuous functions satisfying the same constraints.

To define a topological category, we do exactly the same thing: replace $C_0$ and $C_1$ with spaces, and $i$, $s$, $t$, and $c$ with continuous maps satisfying the same “algebraic relations” outlined above.