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Stream: theory: category theory

Topic: Strange Structure

Joshua Meyers (Oct 03 2020 at 14:25):

Consider a category enriched over the monoidal category $(\text{Set}, \sqcup, \varnothing)$, that is, sets, as usual, but with the coproduct instead of the product as the monoidal operation.

Joshua Meyers (Oct 03 2020 at 14:28):

An example of this structure: Let objects be real numbers, let $\text{Hom}(a,b)=[a,b]=\{x\in\mathbb{R}:a\leq x \leq b\}$, and let composition be $[b,c]\sqcup [a,b] \rightarrow [a,c]$ determined by the inclusions $[b,c]\subseteq [a,c]$ and $[a,b]\subseteq [a,c]$.

Joshua Meyers (Oct 03 2020 at 14:30):

We could also take $\text{Hom}(a,b)$ to be an open interval, a half-open interval, an open interval intersected with the rationals,...

Joshua Meyers (Oct 03 2020 at 14:33):

Usually $\text{Hom}(a,b)$ is taken to mean the collection of ways of going from $a$ to $b$, or the measure (metric space) or possibility (preorder) of going from $a$ to $b$. But here it is taken to mean the collection of things you'll find on the way from $a$ to $b$.

Joshua Meyers (Oct 03 2020 at 14:35):

Does this semantics work with all instances of this structure? Is there a way of understanding this difference in semantics in terms of covariance and contravariance?

Reid Barton (Oct 03 2020 at 14:35):

What happens when $a \le b$ but $c < b$?

Joshua Meyers (Oct 03 2020 at 15:01):

Oh good point, that would mess this up

Reid Barton (Oct 03 2020 at 16:06):

Maybe you could consider enriching in the category $V$ whose objects are maps of sets $A \to X$, with monoidal structure $(A \to X) \otimes (B \to Y) = (A \times Y \amalg X \times B \to X \times Y)$.

Jason Erbele (Oct 03 2020 at 16:24):

Perhaps I am misunderstanding, but even when a ≤ b ≤ c, there seem to be some additional problems with taking Hom(a,b) to be the open intervals (a,b). For instance, Hom(a,a) should include the identity, but the open set (a,a) is empty. Also problematic (unless I am mistaken) is composing two non-empty open intervals, Hom(b,c) with Hom(a,b), seems to give an open set that is no longer an open interval.

Morgan Rogers (he/him) (Oct 03 2020 at 21:17):

Still a fun enrichment to consider, even if this particular example doesn't quite work.

Joshua Meyers (Oct 04 2020 at 00:12):

@Jason Erbele the monoidal unit here is $\varnothing$ so the identity is $\text{id}:\varnothing\rightarrow \text{Hom}(c,c)$, which is uniquely defined even if $\text{Hom}(c,c)$ is empty. Also it doesn't matter that the disjoint union fails to itself be an open interval, the composition is a map from this disjoint union to an open interval.

Joshua Meyers (Oct 04 2020 at 00:13):

So I proved that in any category with this enrichment, all hom-sets must be isomorphic as sets (using the standard definition of isomorphic, not the one relative to this enrichment).

Joshua Meyers (Oct 04 2020 at 00:24):

Let our composition $\text{Hom}(b,c)\sqcup\text{Hom}(a,b)\rightarrow \text{Hom}(a,c)$ be a copairing of $i_{abc}:\text{Hom}(a,b)\rightarrow \text{Hom}(a,c)$ and $j_{abc}:\text{Hom}(b,c)\rightarrow\text{Hom}(a,c)$. Then unitality gives that $i_{abb}$ and $j_{aab}$ are identities; and associativity gives a whole bunch of relations between the $i_{abc}$ and the $j_{abc}$ (you can work it out if you want). One of these relations is $i_{acd} \circ i_{abc}=i_{abd}$. Letting $c=a, d=b$ gives $i_{aab}\circ i_{aba}=i_{abb}$ which is identity, so $i_{aba}$ has a left inverse. Applying this identity again gives $i_{aba}\circ i_{aab} = i_{aaa}$ which is also identity, so $i_{aba}$ has a right inverse. Thus $i_{aba}:\text{Hom}(a,b)\rightarrow\text{Hom}(a,a)$ is an isomorphism. By a dual argument, it can be shown that $\text{Hom}(a,b)$ is isomorphic to $\text{Hom}(b,b)$. Thus all hom-sets are isomorphic.

Joshua Meyers (Oct 04 2020 at 00:29):

@Reid Barton why that enrichment?

Morgan Rogers (he/him) (Oct 04 2020 at 07:49):

Joshua Meyers said:

Let our composition $\text{Hom}(b,c)\sqcup\text{Hom}(a,b)\rightarrow \text{Hom}(a,c)$ be a copairing of $i_{abc}:\text{Hom}(a,b)\rightarrow \text{Hom}(a,c)$ and $j_{abc}:\text{Hom}(b,c)\rightarrow\text{Hom}(a,c)$.

Is it strictly necessary that this be the definition? This seems like a canonical choice, but there could be others, right?

Morgan Rogers (he/him) (Oct 04 2020 at 07:50):

In any case, that's a really cool result!

Joshua Meyers (Oct 04 2020 at 13:46):

It is necessary! Precompose the composition with the inclusions $\text{Hom}(a,b)\subseteq \text{Hom}(a,c)$ and $\text{Hom}(b,c)\subseteq \text{Hom}(a,c)$ to get $i_{abc}$ and $j_{abc}$. Then the universal property of coproduct says that $i_{abc}$ and $j_{abc}$ uniquely determine the composition.

Morgan Rogers (he/him) (Oct 04 2020 at 14:11):

Aha I see now, I was just being slow. :sweat_smile: