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Stream: theory: category theory

Topic: questions about "Carrés exacts et carrés déductifs"

fosco (Jun 30 2021 at 10:12):

So, reading the paper

"Carrés exacts et carrés déductifs." Diagrammes 6 (1981): G1-G17. <http://eudml.org/doc/192983>

I encountered a notion I can't really make sense of. Let me translate in a more modern language the situation Guitart works in.

Definition. An extensor (extenseur) or an extension structure on a category $\mathcal C$ is a function $E : \textsf{Span}(X,Y) \to \mathcal C(X,Y)$ sending a span $(X \xleftarrow{p} A, A \xrightarrow{f} Y)$ to an arrow $E_p(f) : X \to Y$,

The correspondence $E$ subject to the following axioms:

1. For every pair of composable arrows $p,q$ we have $E_{q\circ p}(f)=E_q(E_p(f))$
2. Given a pullback square $\begin{array}{ccc}A &\overset{p'}\to& B \\_{h'}\downarrow && \downarrow_h\\C &\underset{p}\to& D \end{array}$ then $E_p(f)\circ h = E_{p'}(f\circ h')$
3. If we have a diagram $\xleftarrow{p} \underset{h}{\overset{h'}\leftrightarrows} \xrightarrow{q}$ such that $ph'h=p$ and $qhh'=q$, then $E_q(fph')=E_q(E_h(fp))$.

Obscure as these properties may seem, they serve to abstract the properties enjoyed by the left or right extension bifunctor in a 2-category that admits enough structure.

Now, in his "Carrés exacts et carrés déductifs." Guitart claims that whenever $\boldsymbol P$ is a KZ-monad on a 2-category $\mathcal K$, then for every span $f : I \to E, h : I \to J$, and every $\boldsymbol P$-algebra $E$, the arrow

$\text{Ex}_h(f) := J \to \boldsymbol P J \xrightarrow{h^*} \boldsymbol P I \xrightarrow{f_*} \boldsymbol P E \to E$

defines an extension structure (page 7 of the paper), and this sets up a deduction structure (previous page).

I haven't the faintest idea how this can be, for two reasons:

• First, the definition in the paper lacks some information (I have adapted the definition that Guitart gives, without further mention, using -probably?- the correspondent of $h$ in the Kleisli category of $\boldsymbol P$) that I'm not sure I filled up properly.
• Second, and more important, the definition of a deductive structure is given as follows: it is a relation $\vdash$ from the set of finite subsets of a set $E$ to $E$ itself, in such a way that if we denote it as an infix operator $\Gamma \vdash x$, whe $\Gamma$ is a finite subset $\{a_1,\dots, a_n\}$ of $E$ and $x\in E$, a bunch of axioms are satisfied (to the effect that $\vdash$ behaves like the entailment relation for a set of formulas).

Now, however, the deductive structure induced by an extension structure is defined as follows:

image.png

Not only I don't have the faintest idea about how to extend this definition to make sense for a tuple of morphisms to the left of the entailment relation, but I also haven't got any idea of how to prove that $\{f\}\vdash_{\text{Ex}} f$ and that if $\{f\}\vdash_{\text{Ex}} g$ and $\{g\}\vdash_{\text{Ex}} f$, then $f=g$. For the first, does it follows that $\text{Ex}_1 f=f$? And for the second... I'm even more clueless.

Any help?

fosco (Jun 30 2021 at 16:45):

Let me clarify: does it follow from the axioms of an extenseur that $E_1(f)=f$? The only thing I'm able to prove at first sight is that $E_1 = E_1\circ E_1$.

Morgan Rogers (he/him) (Jul 01 2021 at 08:28):

Given any morphism $h$, you can form a pullback square whose vertical morphisms are $h$ and whose horizontal morphisms are identities. Applying 2, we have $E_1(f) \circ h = E_1(f \circ h)$. Taking $f$ to be the identity, we have $E_1(1) \circ h = E_1(h)$, so it suffices to show that $E_1(1) = 1$ for every identity morphism 1.

Amar Hadzihasanovic (Jul 01 2021 at 08:40):

$E_1(1)$ is also idempotent: by (2) with $h := E_1(1)$ we have $E_1(1) \circ E_1(1) = E_1(E_1(1))$, and the latter is equal to $E_1(1)$ by (1).

Amar Hadzihasanovic (Jul 01 2021 at 08:42):

So you just need it to be cancellable, or equivalently you need $E_1(-)$ to be injective...

Morgan Rogers (he/him) (Jul 01 2021 at 08:50):

If $\mathcal{C}$ isn't required to have all pullbacks, I suspect you could construct a monoid counterexample?

Amar Hadzihasanovic (Jul 01 2021 at 09:17):

I think the minimal counterexample may be the monoid with a single idempotent generator, and the $E_{-}(-)$ are all constantly the idempotent element.

Amar Hadzihasanovic (Jul 01 2021 at 09:28):

The fact is that all the equations involve at least one morphism in the image of $E$, and the only one where one of the two sides is not necessarily in the image of $E$ is still of the form $E_x(y) \circ h = E_{x'}(y')$... so if we can let every $E_x(y)$ be some chosen “left-absorbing” morphism of the correct type, then the equations will hold 'trivially'...

Amar Hadzihasanovic (Jul 01 2021 at 09:30):

More precisely, if $\mathcal{C}$ is such that for all objects $X$, we can fix a cocone $c_X$ under $\mathrm{Id}_\mathcal{C}$ with tip $X$, then for any span $(p,f)$ from $Y$ to $X$ we can let $E_p(f)$ be the component $c_X(Y): Y \to X$ of this cocone, and all the equations will hold for trivial reasons.

Amar Hadzihasanovic (Jul 01 2021 at 09:32):

Special case: any monoid with a left absorbing element (a “left zero”).

Notification Bot (Jul 01 2021 at 10:02):

This topic was moved here from #general: co/appreciation > René Guitart by Matteo Capucci (he/him)

fosco (Jul 01 2021 at 19:55):

I agree, yes

fosco (Jul 01 2021 at 19:55):

whoops

fosco (Jul 01 2021 at 19:56):

Amar Hadzihasanovic said:

$E_1(1)$ is also idempotent: by (2) with $h := E_1(1)$ we have $E_1(1) \circ E_1(1) = E_1(E_1(1))$, and the latter is equal to $E_1(1)$ by (1).

I wanted to reply to this