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?

  • $\begingroup$ What does "algebraic relations of a discrete category" mean? For me, a topological category is a category enriched in topological spaces. $\endgroup$ – user97187 Aug 14 '16 at 3:04
  • $\begingroup$ @Ubiquity: I don't understand either... $\endgroup$ – PhysicsMath Aug 14 '16 at 3:10
  • 3
    $\begingroup$ He defines a topological category as a pair $(C_1,C_0)$ with source and target maps and a degeneracy map. The requirement that these satisfy the algebraic relation of "discrete category" imply that this pair (and the maps between them) is actually a category internal to the category of topological spaces, where $C_1$ is the space of morphisms and $C_0$ is the space of objects. $\endgroup$ – Daniel Grady Aug 14 '16 at 4:24
  • 3
    $\begingroup$ "Discrete" is a bit of a distraction here; you might as well say "ordinary category". In other words, as Daniel says, the data of a topological category consists of continuous maps (source and target maps $C_1 \to C_0$, a unit map $C_0 \to C_1$ which gives the identity assignment, and a composition map $C_1 \times_{C_0} C_1 \to C_1$). The equational axioms these data must satisfy are the same as those of ordinary category theory: associativity of composition, the identity axioms, etc. $\endgroup$ – Todd Trimble Aug 14 '16 at 4:36
  • 3
    $\begingroup$ @PhysicsMath The notion of "discrete category" in the sense of the Wikipedia article you linked to is definitely not relevant here. Freed surely just meant "ordinary category" and used the term "discrete category" to emphasize the fact that in "ordinary" categories there is no topology on the objects/morphisms, i.e. they are "discrete sets". It's a slightly unfortunate coincidence of terminology. $\endgroup$ – Dan Ramras Aug 15 '16 at 2:31

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.

  • $\begingroup$ So you agree with Todd that the "discreteness" is a bit distracting and should be deleted? $\endgroup$ – PhysicsMath Aug 14 '16 at 18:08
  • $\begingroup$ Not necessarily. Maybe "ordinary" is clearer than "discrete," but removing the clarification altogether would be confusing. $\endgroup$ – Arun Debray Aug 14 '16 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.