Equivalent conditions to be a lax idempotent contravariant monads

Question

$\require{AMScd}$As already said, a contravariant monad is "like a monad, but a contravariant functor":

• the multiplication and unit are dinatural transformations
• an algebra is a map $a : TA\to A$ such that $$ \begin{CD} TTA @>\mu_A>> TA @. A @>\eta_A>> TA\\ @ATaAA @VVaV @| @VVaV\\ TA @>>a> A @. A @=A \end{CD} $$ are commutative (in whatever sense adapts to the canonical wanna-be example of these structure, the presheaf construction, where moreover $\mu=T\eta$).

I've been persuaded, though, that not all equivalent conditions for $T$ to be a lax-idempotent 2-monad can hold. For example, I don't see why the presheaf construction yields a morphism $b\Rightarrow f . a .Tf$ for a pair of $T$-algebras (=total categories). It is neverthelesee true that these two conditions hold:

1. $a\dashv \eta_A$ for every $T$-algebra $a : TA\to A$.
2. $\mu_A\dashv \eta_{TA}$ for every object $A$.

For a generic covariant lax-idempotent monad, they're equivalent. Here, they both hold for $T=[\,\_^o, Set]$, and it's easy to see that in general 1 implies 2 (simply because free algebras $(TA,\mu_A)$ are algebras); but the proof of the converse implication can't probably be adapted to the case of a contravariant monad.

Is it still true that $1\iff 2$?

Answer 1

It's perhaps not immediately clear what you mean in requiring the unit $\eta$ to be "dinatural", and indeed it won't be dinatural in the strict sense under the only reasonable interpretation I see. It does however carry "lax dinatural" structure. Let's first work through this.

If $F, G: C^{op} \times C \to D$ are two functors, then a dinatural transformation from $F$ to $G$ consists of a family of maps $\theta_c: F(c, c) \to G(c, c)$ such that for all $f: c \to c'$, the hexagon

$$\begin{array}{l} F(c', c) & \stackrel{F(f, c)}{\to} & F(c, c) & \stackrel{\theta_c}{\to} & G(c, c) \\ \downarrow \; F(c', f) & & & & \downarrow \; G(c, f) \\ F(c', c') & \underset{\theta_{c'}}{\to} & G(c', c') & \underset{G(f, c')}{\to} & G(c, c') \end{array}$$

commutes. Now of course $1_C: C \to C$ and $T: C^{op} \to C$ don't have the required form, so we have to massage them into shape: put

$$F = \left(C^{op} \times C \stackrel{\pi_2}{\to} C \stackrel{1_C}{\to} C\right)$$

$$G = \left(C^{op} \times C \stackrel{\pi_1}{\to} C^{op} \stackrel{T}{\to} C\right)$$

Then $F(c, c) = c, G(c, c) = Tc$, and the unit map components $\eta_c: c \to Tc$, if it is to yield a dinatural transformation $(\eta_c: F(c, c) \to G(c, c))_{c \in Ob(C)}$ from $F$ to $G$, would have to render the square

$$\begin{array}{l} c & \stackrel{\eta_c}{\to} & Tc \\ \downarrow\; f & & \uparrow\; Tf \\ c' & \underset{\eta_{c'}}{\to} & Tc' \end{array}$$

commutative. Of course it doesn't in your examples, but there is in the free cocompletion example a lax dinaturality structure

$$\begin{array}{l} c & \stackrel{\eta_c}{\to} & Tc \\ \downarrow\; f & \Downarrow \; \hat{\eta_f} & \uparrow\; Tf \\ c' & \underset{\eta_{c'}}{\to} & Tc' \end{array}$$

In the free cocompletion example, $\eta_c$ is the Yoneda embedding on a category $c$ and this 2-cell $\hat{\eta_f}$ is the evident natural transformation $\hom(-, c) \to \hom(f-, fc)$ attached to a functor $f$. I tend to associate this with a covariant 2-functor $T_!: C \to C$ such that $T_! f \dashv Tf$ for 1-cells $f$; this $T_!$ defines a relative pseudomonad, and an actual pseudomonad if we are working for example in a Yoneda structure in which all 1-cells are admissible. In any case, $\eta: 1_C \to T_!$ will be the unit, a strong natural transformation so that for 1-cells $f: c \to c'$ we have a structural isomorphism $\eta_f: T_! f \circ \eta_c \cong \eta_{c'} \circ f$, and the lax dinaturality constraint is mated to $\eta_f$ via the adjunction $T_! f \dashv Tf$:

$$\frac{T_!f \circ \eta_c \overset{\eta_f}{\to} \eta_{c'} \circ f}{\eta_c \underset{\hat{\eta_f}}{\to} Tf \circ \eta \circ f}$$

Let's consider the (co?)lax dinatural structure on the multiplication side. Morally, in our example, this should again arise through a mating process where a map $f: A \to B$ is a $T_!$-algebra map if there is a coherent isomorphism $\varepsilon_f: f \circ a \cong b \circ T_! f$. This mates to a 2-cell

$$\hat{\varepsilon_f}: f \circ a \circ T f \to b$$

which is different from what you were predicting above. (Morally, a $T_!$-algebra map $f:A \to B$ in our canonical example is a left adjoint, say $f \dashv g: B \to A$, and then the 2-cell $\hat{\varepsilon_f}$ is just the mate via taking left adjoints of the 2-cell $\hat{\eta_g}: \eta_B \to Tg \circ \eta_A \circ g$.) If we take $f = T\eta_c: TTc \to Tc$, then the lax 2-cell for the multiplication takes the form

$$T\eta_c \circ T\eta_{Tc} \circ TT\eta_c \to T\eta_c$$

and the lax constraint in this case turns out to be an isomorphism.

You haven't really said what (you think) a lax idempotent (unidetermined) contramonad is. We know what an ordinary lax idempotent covariant monad is; here $T_!$ plays that role. There are actually a number of details which need to be made precise (and which you haven't made precise in your recent posts). My advice would be to use $T_!$ together with mating procedures as a temporary crutch, carefully translating algebras and lax idempotence of $T_!$ into "$T_!$-free form" (i.e., $T$-only form), but the calculations given above suggest that you may have gotten (or guessed) wrong some of the directions of structural 2-cells. Once you get all this straight, then it seems very plausible that you could prove an analogue of (1) $\Leftrightarrow$ (2).