# Definition of enriched category as lax-monoidal functor


... an alternative way of viewing a $V$-category is as a set $X$ with a (lax) monoidal functor $\Phi=\Phi_d$ of the form $$V^\text{op} \stackrel{yon_V}{\longrightarrow}\mathbf{Set}^V \stackrel{d^*}{\longrightarrow} \mathbf{Set}^{X \times X}$$ where the codomain is identified with the monoidal category of spans on $X$...

In the link above it is shown how from the data of a $V$-enriched category $(X,d \colon X \times X \to V,\comp,\id)$ one can get a lax functor $\Phi$ as above.

Unfortunately it is not clear, at least not to me, how we can reverse the construction, that is how we can get a $V$-enriched category from a lax-monoidal functor $\Phi \colon V^\text{op} \to \mathbf{Set}^{X \times X}$ which factors through the yoneda embedding and $d^*$ for some $d \colon X \times X \to V$.

Indeed for a such $\Phi_d$ I can clearly see that we get from laxness a family of morphisms $$\Phi(v)\odot\Phi(v') \to \Phi(v \otimes v')$$ which, expanding a little bit, amounts to a family of functions $$\coprod_{y \in X}V(v,d(y,z)) \times V(v',d(x,y)) \to V(v \otimes v',d(x,z))$$ but it is not clear how from these data one can retrive a family of mappings $$\comp \colon d(y,z) \otimes d(x,y) \to d(x,z)$$ in $V$.

In the link there is a reference to a yoneda argument which should allow to get the compositions. If we had a family of natural transformations $$\coprod_{y \in X} V(v \otimes v',d(y,z) \otimes d(x,y)) \to V(v \otimes v',d(x,z))$$ then we could clearly retrive the mappings $\comp$ by using yoneda and letting $v'=I$ (the identity of the monoidal category $V$). This situation is verified if the natural transformations $\Phi(v)\odot \Phi(v') \to \Phi(v \otimes v')$ factor through the compositions mappings $$V(v,d(y,z)) \times V(v',d(x,y)) \stackrel{\otimes}\to V(v \otimes v', d(y,z) \otimes d(x,y))$$ but I don't see any reason why this should happen in general for any lax functor of the form $\Phi_d$.

So am I missing something?

Any help will be appreciated.

I only just got wind of this question. The map

$$\comp \colon d(y,z) \otimes d(x,y) \to d(x,z)$$

is obtained from the family of maps

$$\coprod_{y \in X}V(v,d(y,z)) \times V(v',d(x,y)) \to V(v \otimes v',d(x,z))$$

by setting $v = d(y, z), v' = d(x, y)$, and evaluating at the pair of identity maps $(1_{d(y, z)}, 1_{d(x, y)})$.

Please let me know if this doesn't answer the question.

• (Oh by the way: I'm Todd Trimble. (-: ) Jan 10 '18 at 16:16
• Sorry for taking so long, apparently I have missed the notification. To answer your question: yes this answer my question. May 24 '18 at 20:38

It seems that if you read "lax monoidal functor factoring through Yoneda" to suggest that not only the functor, but also the monoidality constraints, factor through those of Yoneda (with its strong structure mapping to the Day product,) you find that your laxness constraints factor through the natural map $\coprod V(v,d())\times V(v',d())\to \coprod V(v\otimes v', d()\otimes d())$, from which you can get the composition in the way you already described. The same, one hopes, may be said for the unit.

• So, you are basically saying that the characterization is requiring that $\Phi_d$ factors in the category of monoidal-categories and lax-monoidal functors, am I right? In particular are you implying that the characterization requires for $d^*$ to be a lax functor too? Nov 26 '16 at 16:00
• Yes, that's right. At the bottom of that nLab paragraph you'll see that they do expect $d^*$ to be (bicontinuous) lax monoidal. I guess that the monoidality is automatic from the universality of the Day product, in any case. Nov 26 '16 at 22:39
• I'm not sure the Day product actually has the universal property I just attributed to it. Nov 26 '16 at 22:46
• I have tried to do the whole computation, but did not get that far. I guess it is because I am not that acquainted with Day convolution. Have you got any reference with enough examples (on the convolution)? Nov 27 '16 at 11:44