Let $\mathcal{C}$ be symmetric monoidal $(\infty,1)$-category, that is, a functor $\mathcal{C} \colon \Delta^{\text{op}} \times \Gamma^\text{op} \to \text{sSet}$, where $\Delta$ is the simplex category, $\Gamma$ is the opposite category of the skeleton of finite pointed sets $\text{Fin}_\star$ (objects are all the sets $\langle l \rangle = \{\star, 1,\ldots, l\}$ and morphisms are basepoint preserving maps), such that $\mathcal{C}$ satisfies Segal's special $\Delta$-condition:
$$\mathcal{C}_{a+b} \overset{\simeq}{\longrightarrow}\mathcal{C}_{a} \times_{\mathcal{C}_0} \mathcal{C}_b$$
Moreover, $\mathcal{C}$ satisfies Segal's special $\Gamma$-condition:
$$\mathcal{C}\langle l \rangle \overset{\prod\limits_{i=1}^l \mathcal{C}\delta_i}{\longrightarrow} (\mathcal{C}\langle 1 \rangle)^l$$
is a weak equivalence, where $\delta_i \colon \langle l \rangle \to\langle 1 \rangle$ is the map which sends every element $j \neq i$ in $\langle l \rangle$ to $\star$ and $i$ to $1$. In particular, $\mathcal{C}$ is fibrant with respect to the injective model structure on the category of simplicial presheaves on $(\Delta\times \Gamma)$ denoted $\text{Psh}_\Delta(\Delta\times \Gamma)$. The $(\infty,1)$-category $\mathcal{C}\langle 1\rangle$ is then called the underlying symmetric monoidal $(\infty,1)$-category. My question is the following:
Given $\mathcal{C}$ as above, we may consider the resulting homotopy category $\mathfrak{h}_1\mathcal{C}\langle 1\rangle$. This category has as its set of objects the elements of the set $\mathcal{C}\langle 1 \rangle_{0,0}$, while for objects $x,y$ the corresponding hom-set is given by
$$\mathfrak{h}_1\mathcal{C}\langle 1 \rangle(x,y) = \pi_0(\{x\} \times_{\mathcal{C}\langle 1\rangle_{0,\bullet}} \mathcal{C}\langle 1 \rangle_{1,\bullet} \times_{\mathcal{C}\langle 1\rangle_{0,\bullet}}\{y\})$$ and composition of morphisms is induced by means of Segal's special $\Delta$-conditions. I want to prove that $\mathfrak{h}_1\mathcal{C}\langle 1 \rangle$ is not just a category, but a symmetric monoidal category.
I have tried the following: I think the idea of how to go about this is pretty clear. We realize that we first need to define a tensor product for the symmetric monoidal structure. This is done by looking at the weak equivalence
$$\mathcal{C}\langle 2 \rangle \overset{\mathcal{C}\delta_1\times \mathcal{C}\delta_2}{\longrightarrow} (\mathcal{C}\langle 1 \rangle)^2$$ In fact, we take a weak inverse of this map and call it $m \colon (\mathcal{C}\langle 1 \rangle)^2 \to \mathcal{C}\langle 2\rangle$ and compose this map with $\mathcal{C}\varphi\colon \mathcal{C}\langle 2\rangle \to \mathcal{C}\langle 1 \rangle$ which is induced by $\varphi \colon \langle 2 \rangle \to \langle 1\rangle, 1,2 \mapsto 1$ to obtain a map $\otimes \colon (\mathcal{C}\langle 1\rangle)^2 \to \mathcal{C}\langle1\rangle$. Passing to the homotopy category, this induces a functor $$\otimes \colon \mathfrak{h}_1\mathcal{C}\langle 1 \rangle \times \mathfrak{h}_1\mathcal{C}\langle 1 \rangle \to \mathfrak{h}_1\mathcal{C}\langle 1 \rangle$$ In order to obtain the braiding isomorphism we shall use the twist map $t \colon \langle 2 \rangle \to \langle 2\rangle, 1,2 \mapsto 2,1$. This gives us an isomorphism $\mathcal{C}(t) \colon \mathcal{C}\langle 2\rangle \to \mathcal{C}\langle 2\rangle$ and the indcued twist map $\tau$ must then be given by the composition
$$\mathcal{C}\langle 1 \rangle \times \mathcal{C}\langle 1 \rangle \overset{m}{\longrightarrow} \mathcal{C}\langle 2\rangle \overset{\mathcal{C}(t)}{\longrightarrow} \mathcal{C}\langle 2\rangle \overset{\mathcal{C}\delta_1\times \mathcal{C}\delta_2}{\longrightarrow} \mathcal{C}\langle 1 \rangle \times \mathcal{C}\langle 1 \rangle$$ We have to show that after having passed to homotopy categories, we have a natural isomorphism
$$\otimes \circ \tau \cong \otimes$$ I am not sure on how to proceed here to obtain this natural isomorphism. In fact, before passing to homotopy categories we should get a weak equivalence $\otimes \circ \tau \simeq \otimes$ of sorts, though I am not sure what I mean by that yet. I think that once one has the correct idea, also the remaining axioms are easy to check.
How to construct the above morphism?
Best Answer
Since $⊗∘τ$ and $⊗$ are both morphisms $\def\cC{{\cal C}}\cC⟨1⟩^2→\cC⟨1⟩$, what we are looking for is a homotopy between them.
We have $$⊗=\cC φ∘m, \qquad τ=(\cC δ_1,\cC δ_2)∘\cC t∘m.$$ Now $$⊗∘τ=\cC φ∘m∘(\cC δ_1,\cC δ_2)∘\cC(t)∘m,$$ which can be simplified by observing that $m∘(\cC δ_1,\cC δ_2)$ is canonically homotopic to identity, given that $m$ has defined as the inverse of $(\cC δ_1,\cC δ_2)$. Thus, $$⊗∘τ≃\cC φ∘\cC t∘m=\cC(φ∘t)∘m=\cC φ∘m=⊗,$$ as desired.