6
$\begingroup$

$\require{AMScd}$Let $\cal K$ be a 2-category, and $j : A\to B$ one of its 1-cells. Assume that the induced map $$ j^* : {\cal K}(B,B)\to {\cal K}(A,B) $$ precomposing with $j$ has a left adjoint $j_!$, the left extension along $j$.

Given 1-cells $x,y\in {\cal K}(A,B)$, it is possible to obtain canonical maps

  • $\gamma_{xyz} : j_! (j_!x\circ y) \circ z \to j_!x\circ j_!y\circ z$, a map that in turn is determined by a map $\bar\gamma : j_!(j_!x\circ y) \to j_!x\circ j_! y$ via the universal property of $j_!$, so that $\gamma = \bar\gamma * z$; in ${\cal K}=\bf Cat$, a sufficient condition for this to be invertible is that left extensions along $j$ are absolute.
  • $\sigma : j_!(j)\to 1_B$ the counit of the density comonad of $j$; in ${\cal K}=\bf Cat$, this is invertible if and only if $j$ is dense.
  • $\eta : x \to j_!(x)\circ j$, the $x$-component of the unit of $j_!\dashv j^*$; in ${\cal K}=\bf Cat$, this is invertible if and only if $j$ is fully faithful.

I want to prove that the following two diagrams of 2-cells are commutative: $$ \begin{CD} j_!(j_!x\circ y)\circ j @>\gamma_{xyj}>> j_!x\circ j_!y\circ j\\ @A\eta_{j_!x\circ y}AA @| \\ j_!(x)\circ y @>>j_!x * \eta_y> j_!x\circ j_!y\circ j \end{CD} \begin{CD} j_!(j_!j\circ x)\circ y @>j_!\sigma * y>> j_!x\circ y \\ @| @AA\sigma * j_!x\circ y A \\ j_!(j_!j\circ x)\circ y @>>\gamma_{jxy}> j_!j\circ j_!x \circ y \end{CD} $$ But, as I expected when I began, this is going to be quite painful. Any advice to make the proof simpler and clearer?

A conceptual proof in ${\cal K}=\bf Cat$, that can be adapted to a generic $\cal K$, is welcome!

Edit (Jul 23, 2018). What I am trying to prove is that the claim that appears as Theorem 3.1 in "Monads need not be endofunctors" holds in a generic 2-category.

$\endgroup$
2
+150
$\begingroup$

There is a fast and down to earth proof. We use the letters $R$ and $L$ to indicate the right and left adjoints of morphisms.

For the first one, just call $a=j_!x\circ y$ and $b=j_!x\circ j_!y$, we have one morphism $\phi:a\to j^*b$ with left adjoint $L\phi:j_!a\to b$. Then your diagram becomes $$\begin{CD} j^*j_!a @>j^*L\phi>> j^*b\\ @A\eta AA@|\\ a@>\phi>>j^*b \end{CD}$$ which is an easy $1$ categorical fact about adjunction. It's just the right adjoint diagram of $L\phi\circ\text{id}=\text{id}\circ L\phi$, with $\eta=R\text{id}$ and $\phi=RL\phi$.

For the second one, observe that you can prove commutativity before composing with $y$: every arrow has the form $\phi\ast y$ for some $\phi$. Now just use adjunction of $j^*$ and $j_!$: your diagram becomes $$\begin{CD} j_!j\circ x @>Rj_!(\sigma\ast\text{id})>> j^*j_!x\\ @|@A\sigma\ast\text{id} AA\\ j_!j\circ x@>\text{id}\ast\eta>>j_!j\circ j^*j_!x \end{CD}$$

We can now unpack $$Rj_!(\sigma\ast\text{id}):j_!j\circ x\to j^*j_!x$$ as $$j_!j\circ x\xrightarrow{\sigma\ast\text{id}} x\xrightarrow{\eta} j^*j_!x$$ To check that this unpacking is right, just observe that $R\eta=\text{id}$. Hence the diagram above becomes

$$\begin{CD} x @>\eta>> j^*j_!x\\ @A\sigma\ast\text{id} AA@A\sigma\ast\text{id} AA\\ j_!j\circ x@>\sigma\ast\eta>>j_!j\circ j^*j_!x \end{CD}$$ which is obvious.

$\endgroup$
  • 1
    $\begingroup$ If someone viewed a previous version of this, it's just because I mistakingly reversed some arrows. With the right direction it's even easier! $\endgroup$ – Giulio Bresciani Jul 23 '18 at 13:51
  • 2
    $\begingroup$ Ah, ho capito qual è l'idea. I was tangled into too many messy diagrams to see this trick. Thx! $\endgroup$ – Fosco Loregian Jul 23 '18 at 14:18
  • 1
    $\begingroup$ Di nulla! An easy proof just had to exist. I don't even know which are the axioms of a skew monoidal category: sometimes having less information is better. $\endgroup$ – Giulio Bresciani Jul 23 '18 at 14:29
  • $\begingroup$ I'm actually surprised nobody noticed such a simple argument before! $\endgroup$ – Fosco Loregian Jul 23 '18 at 15:04
  • 1
    $\begingroup$ "Faccio geometria, ma sono anche categorista, lo faccio per i soldi" :-) $\endgroup$ – Fosco Loregian Jul 23 '18 at 15:17
3
$\begingroup$

(This is not a proof, but just some thoughts on abstracting the construction which could potentially be helpful for finding a more conceptual proof of the coherence laws. In any case, I find it fun to think along these lines...)

I suggest it might be useful to generalize this construction by considering it in bifibrational terms. To understand this generalization, first let's repackage the original 2-category $\mathcal{K}$ by building the lax slice category $\mathcal{K}⫽B$, whose objects are pairs $(A,x : A \to B)$ and whose morphisms $(A_1,x_1) \to (A_2,x_2)$ are pairs $(f : A_1 \to A_2, \alpha : x_1 \to x_2\circ f)$. Assuming $\mathcal{K}$ has all left kan extensions, then the forgetful functor $\mathcal{K}⫽B \to \mathcal{K}$ is a bifibration. Moreover, it has the special property that the fiber over $B$ (= endomorphisms of $B$) is a monoidal category which acts on every other fiber (via postcomposition), and that this monoidal action commutes with the action of pullback (which is defined via precomposition). These are all the things we need in order to define the skew monoidal structure induced by a 1-cell $j$ in $\mathcal{K}$, which we will now consider more abstractly.

Let us assume given:

  1. A bifibration $\mathcal{E} \to \mathcal{B}$ (we write $f_!$ and $f^*$ for the push and pull along any morphism $f$ in $\mathcal{B}$);
  2. An object $b \in B$ (the "target") with a monoidal structure $(\circ,i)$ on the fiber category $\mathcal{E}_b$;
  3. A monoidal action $$* : \mathcal{E}_b \times \mathcal{E}\to \mathcal{E}$$ which is fiber-preserving, in the sense that it lies over the trivial action $1 \times \mathcal{B} \to \mathcal{B}$, and which restricts to the monoidal product on the fiber over $b$;
  4. Finally, observe that by the universal properties of push/pull and the fact that the action $*$ is fiber-preserving, there are canonical maps: \begin{align} f_! (x * y) \to x * f_! y \tag{1} \\ x * f^* y \to f^* (x * y) \tag{2} \end{align} We will assume that (2) is invertible, i.e., that the action preserves cartesian morphisms (but not necessarily opcartesian ones).

Note that all of these axioms are satisfied by the example of $\mathcal{K}⫽B \to \mathcal{K}$ above.

Given any such a structure, I claim that any morphism $j : a \to b$ into the target object induces a skew monoidal structure on the fiber category $\mathcal{E}_a$, where the tensor of two objects $x,y \in \mathcal{E}_a$ is defined by $$x \bullet y := j_! x * y$$ and the unit object is defined by $$I := j^* i.$$

Here is how we build the structure maps:

  • [associator $(x \bullet y) \bullet z \to x \bullet (y \bullet z)$] $$ j_!(j_! x * y) * z \to (j_! x \circ j_! y) * z \cong j_! x * (j_! y * z) $$ where we use (1) above, that $*$ restricts to the monoidal product on the fiber over $b$, and that it is a monoidal action.
  • [left unitor $I \bullet x \to x$] $$ j_! (j^* i) * x \to i * x \cong x $$ where we use the counit of the push-pull adjunction and that $i$ is a unit for the monoidal action.
  • [right unitor $x \to x \bullet I$] $$ x \to j^* j_! x \cong j^* (j_! x \circ i) \to j_! x * j^* i $$ where we use the unit of the push-pull adjunction, that $i$ a right unit for the monoidal product, and the inverse to (2) above.

Now of course we have to check the five coherence laws! I haven't done that, but I suspect it will be more fun than in the original setting of $\mathcal{K}$. What do you think?

$\endgroup$
  • $\begingroup$ Oh, that's precisely the kind of suggestion I was hoping for! I'll let you know if it works asap $\endgroup$ – Fosco Loregian Jul 22 '18 at 23:05
  • 1
    $\begingroup$ Hi! Your fibrational POV is very elegant, it will for sure appear in what I'm writing and properly acknowledged, but Giulio's proof must win as the simplest and clearest :-) $\endgroup$ – Fosco Loregian Jul 23 '18 at 15:06
2
$\begingroup$

For the case $\mathcal{K} = \mathbf{Cat}$, there is indeed a worked out proof that this structure on the functor category $[A,B]$ (given by the tensor product $x \bullet y := j_! x \circ y$ and unit $I := j$) satisfies the five coherence laws of a skew monoidal category, as Theorem 3.1 in:

$\endgroup$
  • 1
    $\begingroup$ (I wouldn't say it is conceptual, but it is very explicit and can probably be adapted to any 2-category.) $\endgroup$ – Noam Zeilberger Jul 21 '18 at 16:23
  • 1
    $\begingroup$ Yes, I know that source (I should have pointed to the paper in the OP). It is in fact precisely what I am trying to generalize, but to no avail until now. $\endgroup$ – Fosco Loregian Jul 21 '18 at 20:33
  • $\begingroup$ To put it shortly, a formal justification of the validity of the axioms (in terms of, say, universal properties of left extensions) is buried in lots of daunting equations... if all else fails I can simply point to the paper, but I don't want to cheat. $\endgroup$ – Fosco Loregian Jul 21 '18 at 20:35
  • $\begingroup$ OK, sorry I didn't ask if you knew this reference (but I think you should indeed revise the question to mention it, since it contains an explicit proof of the result when $\mathcal{K} = \mathbf{Cat}$). There is also a statement of the general 2-categorical case in On monads and warpings, but without proof. $\endgroup$ – Noam Zeilberger Jul 21 '18 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.