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?*