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Stream: learning: questions

Topic: Kelly’s book

Soichiro Fujii (Apr 09 2020 at 14:57):

I have a question about a parenthetical remark on p.36 of Kelly’s book Basic Concepts of Enriched Category Theory. There he discusses universe-enlargement in the enriched setting, as required for the construction of functor category with non-small domain.

Assume $\mathbf{Set}$ is the category of sets in some universe, and $\mathbf{Set}’$ is the category of sets in a larger universe. Typical concrete base categories $\mathcal{V}_0$ for enrichment (such as $\mathcal{V}_0=\mathbf{Cat, Gpd, Ord, Ab,}...$) are categories of models of some theory in $\mathbf{Set}$, so for these we can define an enlarged base category $\mathcal{V}_0’$ as the category of models of the same theory in $\mathbf{Set}’$. The inclusion $\mathbf{Set}\longrightarrow\mathbf{Set}’$ induces the inclusion $\mathcal{V}_0\longrightarrow \mathcal{V}_0’$. Kelly claims that this latter inclusion preserves all limits and colimits existing in $\mathcal{V}_0$. But then he remarks:

(We are supposing that our original universe contains infinite sets; otherwise we do not have the colimit-preservation above.)

I cannot see what kind of problem he had in mind.

John Baez (Apr 09 2020 at 17:54):

Interesting. So he's claiming some colimits change when you pass from a universe of just finite sets to a universe with infinite sets?

Paolo Capriotti (Apr 09 2020 at 19:02):

I'm not sure what theories are allowed here, but for example, taking the theory of domains (integral commutative rings), the functor from finite domains to domains does not preserve products. In fact, since every finite domain is a field, $\mathbb F_2$ is subterminal in finite domains, so $\mathbb F_2 \times \mathbb F_2 ≅ \mathbb F_2$, but in all domains that isomorphism does not hold, since there are two maps $\mathbb F_2[x] → \mathbb F_2$. In fact, that product does not exist in the larger category.

John Baez (Apr 09 2020 at 19:07):

Neat! What's the terminal finite domain? I'm having trouble seeing why $\mathbb{F}_2$ is subterminal.

Paolo Capriotti (Apr 09 2020 at 19:11):

I mean there is at most one morphism $A \to \mathbb F_2$, because $A$ must be a (finite) field, so there is exactly one morphism if $A = \mathbb F_2$, and none otherwise.

Morgan Rogers (he/him) (Apr 09 2020 at 19:13):

That's a property of subterminal objects in a category that has a terminal object, but it seems like a misnomer when there isn't one :joy:

John Baez (Apr 09 2020 at 19:20):

Okay, Morgan's remark explains my confusion. For me, a subterminal object is a subobject of a terminal object. But I can imagine someone defining it as an object x where hom(-,x) is always a subterminal set.

Morgan Rogers (he/him) (Apr 09 2020 at 20:02):

Related to the problem of expanding the universe, not covered by the assumption that there is an infinite set: I could express a negation of the axiom of choice as the statement that there is some infinite product of sets which is empty. If I manage to find/construct a universe of sets that satisfies this axiom, there seems to be no reason why the inclusion of this universe into a larger one should preserve this product.

John Baez (Apr 09 2020 at 20:19):

Is "this axiom" the axiom of choice or its negation? (By "converse" you really mean "negation".)

Soichiro Fujii (Apr 10 2020 at 01:32):

Thanks a lot for the interesting comments! There is neither a precise definition of “theory” in Kelly’s book, nor of “universe”. Also $\mathcal{V}_0$ is assumed to be at least complete and cocomplete (i.e., having all $\mathbf{Set}$-small limits and colimits), and for that matter, to have a symmetric monoidal closed structure.

Morgan Rogers (he/him) (Apr 10 2020 at 09:51):

John Baez said:

Is "this axiom" the axiom of choice or its negation? (By "converse" you really mean "negation".)

I do indeed. :+1: I shall correct my comment.

John Baez (Apr 10 2020 at 18:40):

Again: when you say "If I manage to find/construct a universe of sets that satisfies this axiom", does "this axiom" mean the axiom of choice or its negation?

John Baez (Apr 10 2020 at 18:42):

I guess you must mean the axiom of choice, since you're talking about a product that exists in some universe, and wondering if it still exists in a larger universe.

John Baez (Apr 10 2020 at 18:43):

I believe a product in some universe will remain a product in some larger universe, but I'm not sure. Set theorists would call this a question about upwards absoluteness.

Min Ro (Apr 10 2020 at 20:54):

I hope people don't mind a dumb question: If I wanted to consider metric (or enriching over Lawvere-type metric spaces) enriched categories, would Kelly's book be useful for this?

John Baez (Apr 10 2020 at 20:58):

Yes, if you are willing to learn much more than you want to know.

Min Ro (Apr 10 2020 at 22:29):

That's fine. My concern was that I would get through a substantial bit and realize that it doesn't apply to my situation at all.

John Baez (Apr 10 2020 at 22:48):

It all applies. But this book is famous for its high level of generality and abstraction. I haven't penetrated most of it yet.

Joe Moeller (Apr 11 2020 at 00:44):

Min Ro said:

I hope people don't mind a dumb question: If I wanted to consider metric (or enriching over Lawvere-type metric spaces) enriched categories, would Kelly's book be useful for this?

I made the description of this stream "no question is dumb" for a reason.

Faré (Apr 11 2020 at 05:18):

In my work on implementations, I rely a lot on spans where one leg is a full embedding—thus modelling "partial functors". Is there extent work on such spans? Maybe to be called "catafunctors" by opposition to "anafunctors" where on leg is surjective?

Faré (Apr 11 2020 at 05:21):

Other stupid question—my work also uses diagrams that mix "horizontal" morphisms (state transitions) with "vertical" application of a functor—am I doing it wrong? e.g. https://imgur.com/gallery/YrlN9sV

Morgan Rogers (he/him) (Apr 11 2020 at 09:17):

John Baez said:

Again: when you say "If I manage to find/construct a universe of sets that satisfies this axiom", does "this axiom" mean the axiom of choice or its negation?

I meant the 'naive' negation I stated, which says "there is an infinite product of non-empty sets which exists but is empty", as opposed to the axiom of choice which says "I can find an element in any infinite product of non-empty sets".

Morgan Rogers (he/him) (Apr 11 2020 at 10:30):

Actually I have an example. Consider the actions of $\mathbb{Z}$ on finite sets, equipped with the obvious homomorphisms. This is an elementary topos: it's cartesian closed, with products inherited from $\mathrm{Set}$ and $[X,Y] = \mathrm{Hom}(\mathbb{Z} \times X,Y)$, on which $\mathbb{Z}$ acts by composition in the first argument. The trivial action of $\mathbb{Z}$ on the two-element set gives the subobject classifier. If I take the cyclic groups of prime order as $\mathbb{Z}$-sets, the infinite product of all of them in this category is the empty $\mathbb{Z}$-set.

Morgan Rogers (he/him) (Apr 11 2020 at 10:36):

Now for a larger universe of sets than the finite ones in which countable products are defined in the way we would expect, I have a full, faithful, logical functor from the topos described above to the category of actions of $\mathbb{Z}$ on sets in the larger universe, which will also form an elementary topos. Therefore if we instead take these toposes of $\mathbb{Z}$-sets as our universes of sets, I have described an inclusion of universes which does not preserve countable products, since the product of the prime-order $\mathbb{Z}$-sets in the larger universe will not be empty.

Morgan Rogers (he/him) (Apr 11 2020 at 11:01):

(Side note: it's at times like this that I am grateful for all of the work that went into obtaining a neat axiomatisation of elementary toposes. I was able to come up with an example in a few lines thanks to that, and even if I've made an error somewhere it should be easy for someone to spot. I don't know if I could extend this to a counterexample to Kelly's claim, though, since the former topos doesn't have a natural number/infinite object.)

Paolo Capriotti (Apr 11 2020 at 11:50):

Why is the product in finite $\mathbb Z$-sets empty? I think it just doesn't exist there.

Morgan Rogers (he/him) (Apr 11 2020 at 11:56):

Paolo Capriotti said:

Why is the product in finite $\mathbb Z$-sets empty? I think it just doesn't exist there.

The empty set admits morphisms to all of them, and is the only $\mathbb Z$-set which does up to isomorphism, so it's universal.

Paolo Capriotti (Apr 11 2020 at 12:02):

Ah, of course! I forgot that $\mathbb Z$ is not finite :blush:. Nice example!

John Baez (Apr 11 2020 at 16:30):

Okay, that's a very nice example, @Morgan Rogers. Now I get what you're talking about.

I don't know if there are "universes" in the sense of ZF set theory that 1) contain infinite sets, 2) don't obey the axiom of choice but 3) embed in larger universes that do obey the axiom of choice. I should ask some logicians.

Gershom (Apr 11 2020 at 19:37):

If I have a topos that models "normal ZF sans-C" and take its booleanization, as one does in topos-theoretic forcing, does that count, or does that not give you an embedding in the appropriate sense?

John Baez (Apr 12 2020 at 01:03):

I was thinking about an embedding in the usual sense of model theory in classical logic, nothing about topoi. But I don't think I understand what I'm talking about well enough to proceed without a lot more thinking.

zigzag (Apr 14 2020 at 20:31):

If you are only talking about model-theoretic embedding, then yes there are plenty of those exemples. The way set theorists like to build permutation models (which are typically used to break choice) is by first building a forcing extension $\mathcal{M}[G]$ of some model $\mathcal{M}$ based on some poset $P$, and then pick a filter $\mathcal{F}$ on the group of automorphisms of $P$ which allow to define a transitive submodel $\mathcal{S}$ of $\mathcal{M}[G]$ of the so called hereditarily symmetric elements of $\mathcal{M}[G]$ (essentially, a set of $x \in \mathcal{M}[G]$ is symmetric iff the set of automorphisms $\{ \pi \mathrel{|} \pi(x) = x \}$ falls in the chosen filter $\mathcal{F}$). The construction of Cohen can be regarded as a particular case of this.

So if you have $\mathcal{M}$ satisfying ZFC, you get the embedding $\mathcal{S} \subseteq \mathcal{M}[G]$ that satisfies 1-3.