# Join of simplicial categories

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 $\mathrm{Map}_{\mathcal{E}^{\triangleright}}(*,*) := \Delta^0$.

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).
• 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 $\mathcal{C} \star' \mathcal{D} \rightarrow \mathcal{E}$. – Daniel Gerigk Oct 25 '15 at 20:56
• (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.) – Daniel Gerigk Oct 25 '15 at 21:13
• 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.) – Omar Antolín-Camarena Oct 26 '15 at 18:17
• 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. – Daniel Gerigk Oct 26 '15 at 18:43
• 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..." – Omar Antolín-Camarena Oct 26 '15 at 19:04