2
$\begingroup$

Let $\mathcal{C},\mathcal{D}$ be simplicial categories.

Of course, we have the "naïve" join $\mathcal{C} \star \mathcal{D}$, which has $$ \mathrm{Ob}(\mathcal{C} \star \mathcal{D}) := \mathrm{Ob}(\mathcal{C}) \sqcup \mathrm{Ob}(\mathcal{D}) $$ and $$ \mathrm{Map}_{\mathcal{C} \star \mathcal{D}}(x,y) := \begin{cases} \mathrm{Map}_{\mathcal{C}}(x,y), & \text{if $x,y \in \mathcal{C}$}.\\ \mathrm{Map}_{\mathcal{D}}(x,y), & \text{if $x,y \in \mathcal{D}$}.\\ \Delta^0, & \text{if $x \in \mathcal{C}$ and $y \in \mathcal{D}$}.\\ \emptyset, & \text{if $x \in \mathcal{D}$ and $y \in \mathcal{C}$}. \end{cases} $$

But that is not the kind of join that I want. I would like to have the join $\mathcal{C} \star' \mathcal{D}$ be defined as the simplicial category which is generated by the conditions that $$ \mathrm{Ob}(\mathcal{C} \star' \mathcal{D}) = \mathrm{Ob}(\mathcal{C}) \sqcup \mathrm{Ob}(\mathcal{D}), $$ that $$ \mathrm{Map}_{\mathcal{C} \star' \mathcal{D}}(x,y) = \begin{cases} \mathrm{Map}_{\mathcal{C}}(x,y), & \text{if $x,y \in \mathcal{C}$},\\ \mathrm{Map}_{\mathcal{D}}(x,y), & \text{if $x,y \in \mathcal{D}$}, \end{cases} $$ and that, for $x \in \mathcal{C}$ and $y \in \mathcal{D}$, the mapping space $\mathrm{Map}_{\mathcal{C} \star' \mathcal{D}}(x,y)$ has a strongly initial object (in the sense of Definition 1.2.12.3 of Lurie's Higher Topos Theory).

Does a simplicial category $\mathcal{C} \star' \mathcal{D}$ having this universal property exist? And if so, where can I read more about this version of the join of two simplicial categories?

EDIT - 20.10.2015

A motivation for my question is that I think that $\mathfrak{C}\left[\Delta^n\right]$, the Cordier straightening of $\Delta^n$, is generated (as a simplicial category) by the conditions that $$\mathrm{Ob}(\mathfrak{C}\left[\Delta^n\right]) = \{0,1,\dots,n\}$$ and that for every $0 \leq i < j \leq n$, the mapping space $\mathrm{Map}_{\mathfrak{C}\left[\Delta^n\right]}(i,j)$ has a strongly initial object. If that - and the above construction of the join - is correct, then we should have $$\mathfrak{C}\left[(\Delta^n)^{\triangleright}\right] = \mathfrak{C}\left[\Delta^n \star \Delta^0\right] = \mathfrak{C}\left[\Delta^n\right] \star' \mathfrak{C}\left[\Delta^0\right].$$

EDIT - 31.10.2015

I think I have found a nice construction of the join $\mathcal{C} \star' \mathcal{D}$: we have $$\mathcal{C} \star' \mathcal{D} = \mathcal{C}^{\triangleright} \cup_{\{*\}} \mathcal{D}^{\triangleleft},$$ where $\{*\}$ denotes the vertex of both the cone $\mathcal{C}^{\triangleright}$ and the cone $\mathcal{D}^{\triangleleft}$ (which are defined below).

Let $\mathcal{E}$ be a simplicial category. We will only describe the cone $\mathcal{E}^{\triangleright}$ - the cone $\mathcal{E}^{\triangleleft}$ is defined analogously. We define $\mathrm{Ob}(\mathcal{E}^{\triangleright}) := \mathrm{Ob}(\mathcal{E}) \sqcup \{*\}$ and stipulate that $\mathcal{E}$ is a full simplicial subcategory of $\mathcal{E}^{\triangleright}$. Let $x \in \mathcal{E}$. We define the $n$-simplices of the mapping space $\mathrm{Map}_{\mathcal{E}^{\triangleright}}(x,*)$ to be the set of morphisms $\mathfrak{C}\left[\Delta^{n+1}\right] \rightarrow \mathcal{E}$ that send the object $0$ to $x$. For a morphism $\alpha : \Delta^m \rightarrow \Delta^n$, the map $(\mathrm{Map}_{\mathcal{E}^{\triangleright}}(x,*))_n \rightarrow (\mathrm{Map}_{\mathcal{E}^{\triangleright}}(x,*))_m$ is induced by the map $\mathfrak{C}\left[\Delta^{m+1}\right] \rightarrow \mathfrak{C}\left[\Delta^{n+1}\right]$ which is induced by the map $\Delta^{m+1} = \Delta^0 \star \Delta^m \rightarrow \Delta^0 \star \Delta^n = \Delta^{n+1}$. The composition $$\mathrm{Map}_{\mathcal{E}^{\triangleright}}(x,*) \times \mathrm{Map}_{\mathcal{E}^{\triangleright}}(y,x) \rightarrow \mathrm{Map}_{\mathcal{E}^{\triangleright}}(y,*)$$ for $y \in \mathcal{E}$ is more complicated, but shouldn't cause too much trouble. Finally, we set $\mathrm{Map}_{\mathcal{E}^{\triangleright}}(*,y) := \emptyset$ for all $y \in \mathcal{E}$ and $\mathrm{Map}_{\mathcal{E}^{\triangleright}}(*,*) := \Delta^0$.

Can you confirm that this construction works? Where can I find more information about this kind of join of two simplicial categories?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

EDIT: I had misunderstood the question when writing the answer below. I thought Daniel wanted a simplicial category with the given conditions and pointed out the usual join satisfies them, but Daniel wants an initial such category.

The first definition you gave has the property you ask for: the unique object of $\Delta^0$ is strongly initial (the name 'strongly initial' would be awful if the object of the terminal category weren't an example).

Improve this answer
$\endgroup$
5
  • $\begingroup$ Yes, sure, but the "naïve" join $\mathcal{C} \star \mathcal{D}$ is not "generated" by those conditions; by that, I mean that $\mathcal{C} \star' \mathcal{D}$ shall fulfill those conditions, and for every simplicial category $\mathcal{E}$ fulfilling those conditions, there shall be a unique functor $\mathcal{C} \star' \mathcal{D} \rightarrow \mathcal{E}$. $\endgroup$
    – Daniel Gerigk
    Oct 25, 2015 at 20:56
  • $\begingroup$ (I forget to mention: that unique functor $\mathcal{C} \star' \mathcal{D} \rightarrow \mathcal{E}$ shall, of course, "respect" the simplicial subcategories $\mathcal{C}$ and $\mathcal{D}$ and the stipulated strongly initial objects.) $\endgroup$
    – Daniel Gerigk
    Oct 25, 2015 at 21:13
  • $\begingroup$ Are you sure $\Delta^0$ isn't what you want? If you have an $\mathcal{E}$ satisfying the conditions you ask and with chosen compatible strongly initial objects, isn't there a unique functor $\mathcal{C} \star \mathcal{D}$ defined as the identity on the morphisms from $\mathcal{C}$ or $\mathcal{D}$ and sending the $\Delta^0$'s to the chosen strongly initial objects? (And if by "stipulated" you don't mean chosen I don't think you have that much hope of the functor being unique --only unique up to homotopy.) $\endgroup$
    – Omar Antolín-Camarena
    Oct 26, 2015 at 18:17
  • $\begingroup$ Consider, for example, the case $\mathcal{C} := \mathfrak{C}\left[\Delta^1\right]$ and $\mathcal{D} := \mathfrak{C}\left[\Delta^0\right]$. Then $\mathcal{C} \star' \mathcal{D} = \mathfrak{C}\left[\Delta^2\right]$, and there is no functor $\mathcal{C} \star \mathcal{D} \rightarrow \mathfrak{C}\left[\Delta^2\right]$ which restricts to the identity on the simplicial subcategories $\mathcal{C}$ and $\mathcal{D}$ and sends chosen strongly initial objects to chosen strongly initial objects. $\endgroup$
    – Daniel Gerigk
    Oct 26, 2015 at 18:43
  • $\begingroup$ Oh, I see, in $\mathfrak{C}\left[\Delta^2\right]$ there are unique initial objects but they are not "compatible", so that there is no functor from $0 \to 1 \to 2$ that picks out the initial objects. It's not clear to me that this "generated" business works, i.e., I don't know if there really is an initial $\mathcal{E}$ with strongly initial elements in the hom spaces from objects of $\mathcal{C}$ to $\mathcal{D}$. If you have a construction you might add it to the question, if not you could modify the question to "does there exist a $\mathcal{C} \star' \mathcal{D}$ such that..." $\endgroup$
    – Omar Antolín-Camarena
    Oct 26, 2015 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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