The category of semisimplicial sets has the structure of a monoidal category by the geometric product $\otimes$, see for example Rourke and Sanderson's paper '$\Delta$-sets I: Homotopy Theory'. This geometric product has the property that for the left adjoint $L$ of the forgetful functor $U$ from simplicial sets to semisimplicial sets, we have $L(X\otimes Y)\cong L(X)\times L(Y)$. Moreover, it is discussed in Sattlers paper on constructive homotopy theory that the forgetful functor preserves Kan complexes, and that the right adjoint $R$ of the forgetful functor preserves Kan complexes. I was hoping to deduce from this that given two semisimplicial Kan complexes $X$ and $Y$, their geometric product $X\otimes Y$ would also be a tensor product, but I failed to do so. Does anyone have a proof, reference or counterexample?


Thanks!



Does the monoidal structure on semisimplicial sets preserve fibrant objects?
It is not the case: the terminal semi-simplicial set $1$ is obviously fibrant but as I will show below the geometric product $1 \otimes 1$ is not fibrant.


**1) What does $1 \otimes 1$ look like ?**


So, $1 \otimes 1$ identifies with the subset of non-degenerate cells of $L(1 \otimes 1) = L 1 \times L 1$.


In general $LX$ admits an explicit desciprion as:


$$ (LX)\_n = \{ s: [n] \twoheadrightarrow [k], x \in X\_k \} $$
with the functoriality being given using the image factorization in $\Delta$.


In particular, $(L1)\_n$ coincide with the set of surjection $[n] \twoheadrightarrow [k]$ in $\Delta$.


A cell of $L1 \times L1$ is hence a pair of such surjection $[n] \twoheadrightarrow [k]$ and $[n] \twoheadrightarrow [k']$. Again, the functoriality is given by precomposition and image factorization on each component. So a cell is non-degenrate if and only if the joint map $[n] \to [k] \times [k']$ is injective.


Hence, $1 \otimes 1$ has for $n$-cells the injective maps $[n] \to [k] \times [k']$ such that each component taken separately is surjective. The functoriality is given by precomposing by a map $[n'] \to [n]$ and dropping from $[k]$ and $[k']$ the element that are no longer in the image.


**2) Why is it not fibrant ?**


I claim that $1 \otimes 1$ does not have the lifting property against the semi-simplicial horn inclusion $\Lambda\_+^0[2] \to \Delta\_+[2]$.


$1 \otimes 1$ has only one $0$-cells given by the unique map $([0] \to [0] \times [0])$. SO any pair of $1$ cell give us a map $\Lambda\_+^0[2] \to 1 \otimes 1$. We take the one that sends the side $\{0,2\}$ to the unique isomorphism $[1] \to [1] \times [0]$ and the side $\{0,1\}$ to the diagonal map $[1] \to [1] \times [1]$.


I claim this horn cannot be filled. Indeed assume it has a filler $v$. That would be a map $v:[2] \to [k] \times [k']$ such that the restriction of $v$ to $\{0,2\}$ and $\{0,1\}$ are, once we drop from $[k]$ and $[k']$ the element that are no longer in the image, the two map we chose above.


Looking at the restriction to $\{0,2\}$, it should be the map $[1] \to [1] \times [0]$, but this forces the second component of $v(0)$ and $v(2)$ to be the same.


But if look at the restriction of $v$ to $\{0,1\}$, you need to have the second component of $v(0)$ and $v(1)$ to be distinct, so this is a contradiction.


