$\mathfrak{h}(\mathcal{G}^{\Delta^1})$ is the category of commutative squares in $\mathfrak{h}\mathcal{G}$

57 Views Asked by At

This is related to the question Homotopy category of a symmetric monoidal $(\infty,1)$-category is symmetric monoidal.

Let us fix an $(\infty,0)$-category $\mathcal{G}$ (a Kan complex). We want to look at the homotopy category $\mathfrak{h}(\mathcal{G}^{\Delta^1})$, which has as its set of objects all natural transformations $\Delta^1 \to \mathcal{G}$ (by the Yoneda Lemma elements in $\mathcal{G}_1$) and as its set of morphisms the natural transformations $\Delta^1 \times \Delta^1 \to \mathcal{G}$. A natural transformation $\eta \colon \Delta^1 \times \Delta^1 \to \mathcal{G}$ breaks down to the information of a square with faces $\eta(\text{id}, d^1), \eta(\text{id}, d^0), \eta(d^1, \text{id}), \eta(d^0, \text{id})$ depicted by the diagram. I want to prove that by passing to the homotopy category $\mathfrak{h}\mathcal{G}$ any such diagram yields a commutative square. The claim would follow, if we had a diagonal morphism (or two homotopical diagonal morphisms) yielding two triangles, such that each given triangle may also be filled by a $2$-simplex. Pictorially, diagram2. Of course, by means of horn filling conditions we may always obtain such a diagram, however, it is not clear (at least to me) why any two such diagonals for the respective triangles must coincide up to homotopy. So:

How to prove that morphisms in $\mathfrak{h}(\mathcal{G}^{\Delta^1})$ are really just commutative squares in $\mathfrak{h}\mathcal{G}$, or put differently, how to prove that we have an isomorphism of categories $\mathfrak{h}(\mathcal{G}^{\Delta^1}) \cong \mathfrak{h}\mathcal{G}^{\mathfrak{h}\Delta^1}$ (note that $\mathfrak{h}\Delta^1$ is precisely the category with two objects and one non-identity morphism)?

Edit: If $\mathcal{G}$ is a Kan complex, then we define $$\mathfrak{h}\mathcal{G}$$ to be the category which has as its set of objects $\mathcal{G}_0$ and as its set of morphisms $\mathcal{G}_1$ modulo the relation below. Domain and codomain maps are given by $\text{dom}= \mathcal{G}(d^1), \text{cod} = \mathcal{G}(d^0)$ and composition is induced by the horn filling properties of $\mathcal{G}$, that is, if $f \colon x \to y,g \colon y \to z$ are in $\mathcal{G}_1$, then there exists a $2$-simplex $\sigma \in \mathcal{G}_2$ so that $d_0(\sigma) = g, d_2(\sigma) = f$ and we define $gf = d_1(\sigma)$. More precisely, $\mathfrak{h} \colon \text{sSet} \to \text{Cat}$ is left adjoint to the usual nerve functor.