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Stream: learning: questions

Topic: Adjunctions Internal to Prof / Dualizable Profunctors

Joshua Meyers (Oct 03 2023 at 18:18):

Hi,

I'm wondering what sort of profunctors $M:C^{\text{op}}\times D \to \text{Set}$ and $N:D^{\text{op}}\times C \to \text{Set}$ form an adjunction internal to $\text{Prof}$. By this I mean that there are natural transformations $\eta:\text{Hom}_C\Rightarrow M\otimes N$ and $\epsilon: N\otimes M \Rightarrow \text{Hom}_D$ satisfying the triangle identities.

nLab has a enigmatic remark about "semifunctors" in this situation but I'm not sure how a semifunctor would specify a profunctor.

In my own thinking, so far I have concluded that for all $c\in C$, $M(c,-)$ and $N(-,c)$ must be quotients of representables, but that's all I have so far.

Joshua Meyers (Oct 03 2023 at 18:26):

Here is the argument for the last claim:

For $f:c\to c'$, denote $\eta_{c,c'}(f):=\overline{(\eta_1(f),\eta_2(f))}$.

Consider one of the triangle identities: $(M\cong \text{Hom}_C\otimes M\xRightarrow{\eta\otimes M} M \otimes N\otimes M \xRightarrow{M\otimes \epsilon} M\otimes\text{Hom}_D\cong M) = id_M$.

Take any $m:M(c,d)$ and run it through this identity:

$m\mapsto \overline{(\text{id}_c,m)} \mapsto \overline{(\eta_1(\text{id}_c), \eta_2(\text{id}_c), m)} \mapsto \overline{(\eta_1(\text{id}_c), \epsilon(\overline{\eta_2(\text{id}_c),m}))}\mapsto M(\text{id}_c, \epsilon(\overline{\eta_2(\text{id}_c),m}))(\eta_1(\text{id}_c))$

By the identity, we must have $m = M(\text{id}_c, \epsilon(\overline{\eta_2(\text{id}_c),m}))(\eta_1(\text{id}_c))$. So then every $m:M(c,d)$ is the result of a $D$-action of the single element $\eta_1(\text{id}_c)$, which only depends on $c$. Thus there is a surjective transformation $D(d,-)\Rightarrow M(c,-)$, where $d$ is the codomain of $\eta_1(\text{id}_c)$, sending $h:d\to d'$ to $M(\text{id}_c, h)(\eta_1(\text{id}_c))$.

A similar argument can be made for $N(-, c)$.

Mike Shulman (Oct 03 2023 at 19:11):

Talking about semifunctors is just a cute trick for talking about functors to the Cauchy completion. If $F:C\to D$ is a semifunctor, then the image of each identity $F(\mathrm{id}_c)$ is not necessarily an identity, but it is idempotent, and thus $F$ induces an actual functor into the Cauchy completion of $D$. And oppositely.

Joshua Meyers (Oct 03 2023 at 19:24):

So the Cauchy completion $D\to\overline{D}$ sends all idempotents to identity? I thought it just makes all idempotents split.

But how does any of this give a profunctor?

Mike Shulman (Oct 03 2023 at 20:15):

I didn't mean that the composite $C \xrightarrow{F} D \to \overline{D}$ is a functor, only that from $F$ you can define a functor $C\to \overline{D}$, by sending an object $c\in C$ to the splitting of the idempotent $F(\mathrm{id}_c)$.

Mike Shulman (Oct 03 2023 at 20:17):

Once you have such a functor, you get a profunctor $C\times \overline{D}^{\mathrm{op}}\to\mathrm{Set}$ in the usual way, and then restrict to $C\times D^{\mathrm{op}}$.

Mike Shulman (Oct 03 2023 at 20:18):

Since $\mathrm{Set}$ has split idempotents, functors $C\times \overline{D}^{\mathrm{op}}\to\mathrm{Set}$ are equivalent to functors $C\times {D}^{\mathrm{op}}\to\mathrm{Set}$, so the restriction doesn't lose any information.