8
$\begingroup$

The following is an excerpt from Lawvere's Some thoughts on the future of category theory.

To clarify the above considerations, generalize to distributive categories and seek philosophical guidance. Even though the determination of which maps are epimorphisms is the more profound question studied with Grothendieck topologies, it takes place within a topos of the following kind. Call a small category $\mathsf C$ "extensive" if it has finite coproducts which yield an equivalence $\mathsf C/A+B = \mathsf C/A \times \mathsf C/B$ and $\mathsf C/ \mathbf{0} = \mathbf{1}$ (this seems a minimum requirement on an op-fibration to conform with the notion of "family" and with Grassmann's "combinatorics of continuous magnitudes"); for example. "the" homotopy category or the category of spaces of dimension at most 4.

  1. In what sense does extensivity make the codomain opfibration conform with the notion of family? Why is it worthy of being called minimal?
  2. What is Grassman's "combinatorics of continuous magnitudes"? Where can I read about it? How and why does extensivity furnish/oil it?
Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

The only way to know for sure is to ask Lawvere. You could try mailing him.

For (1), I guess what he had in mind is the following.

A family $(X_i)_{i \in I}$ of sets indexed by a set $I$ is essentially the same thing as a single set $X$ together with a map $\pi: X \to I$. The idea here is that $X = \coprod_{i \in I} X_i$ and that $\pi$ sends an element $x \in X_i$ to $i$. Once you've also done some thinking about maps between families, you conclude that the category of $I$-indexed families of sets is equivalent to the slice category $\mathbf{Set}/I$.

Now, for any two sets $I$ and $J$, a family indexed over $I + J$ should consist of a family indexed over $I$ together with a family indexed over $J$. That's a theorem if we're talking about families of sets: $$ \mathbf{Set}/(I + J) \simeq \mathbf{Set}/I \times \mathbf{Set}/J. $$ But if we're attempting to generalize the notion of family to categories $\mathbf{C}$ other than $\mathbf{Set}$ (as Lawvere is doing), then this identity becomes a sensible axiom. Thus, he requires that $$ \mathbf{C}/(A + B) \simeq \mathbf{C}/A \times \mathbf{C}/B $$ for all $A, B \in \mathbf{C}$.

Similarly, but more trivially, it's a theorem that there's exactly one family of sets indexed over the empty set: $$ \mathbf{Set}/\emptyset \simeq \mathbf{1}. $$ Hence Lawvere imposes the axiom $$ \mathbf{C}/0 \simeq \mathbf{1} $$ on $\mathbf{C}$.

That's an explanation of why those two axioms on $\mathbf{C}$ are reasonable if you want to be able to talk about "families" in $\mathbf{C}$ and have them behave at all like families of sets. And though one can imagine imposing further axioms on $\mathbf{C}$ to make families in $\mathbf{C}$ behave even more like families of sets, you don't have to: hence, "minimal".

I have no idea about the Grassmann thing. A short web search suggests that when Lawvere puts the words "combinatorics of continuous magnitudes" in quotation marks, he's not actually quoting anyone, but indicating that he's coining a term. It's a shame he doesn't give a reference.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Once you've assumed the axiom about slices over sums of two objects $A+B$ and noticed that it implies the analogous result about slices over sums of $n$ objects for all $n\geq1$ (by induction on $n$, it is practically a reflex action to also assume the result for $n=0$, i.e., the axiom for the slice over $0$. $\endgroup$
    – Andreas Blass
    Commented Sep 9, 2017 at 18:55
  • $\begingroup$ Absolutely agreed. I've preached that sermon so many times that I couldn't bring myself to make the same point once again here. But even with my undergrad students, I try to drum in the lesson that an empty sum should be evaluated as zero---how much would they expect to be charged in a supermarket if they bought no things? And once you've got that idea, it's a short step to understanding that the empty product should be evaluated as 1, the empty union as $\emptyset$, etc. $\endgroup$
    – Tom Leinster
    Commented Sep 10, 2017 at 0:32
  • 1
    $\begingroup$ While I agree of course, it's a funny fact that in the case of extensivity, the binary condition actually also implies the nullary one; see Carboni-Lack-Walters, "Introduction to extensive and distributive categories", Prop 2.8. $\endgroup$
    – Mike Shulman
    Commented Sep 10, 2017 at 7:14
  • $\begingroup$ @TomLeinster where can I read about $\mathbf{Set}/I\simeq [I,\mathsf{Set}]$? $\endgroup$
    – Arrow
    Commented Sep 26, 2017 at 8:31
  • $\begingroup$ There's a bit about it on p.190 of Mac Lane and Moerdijk's book "Sheaves in Geometry and Logic". But they don't say much more than what's at the start of my answer. It must be discussed in lots of other places, though. $\endgroup$
    – Tom Leinster
    Commented Sep 26, 2017 at 20:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .