Why does the tangent classifier classify the tangent (micro)bundle? Let $\mathcal{M}\mathrm{fld}_n$ denote the $\infty$-category of topological manifolds (without boundary) and embeddings; more precisely, it is the homotopy coherent nerve of the simplicial category whose objects are topological manifolds, and whose hom-spaces are given by $\operatorname{Sing}\operatorname{Emb}(M,N)$, where $\mathrm{Emb}(M,N)$ is equipped with the compact-open topology. In Factorization homology of topological manifolds by Ayala-Francis, they construct a tangent classifier functor $$\tau: \mathcal{M}\mathrm{fld}_n\to\mathcal{S}_{/B\mathrm{Top}(n)}$$
which maps a manifold $M$ to the classifying map $M\to B\mathrm{Top} (n)$ of the tangent microbundle of $M$. I do not understand this construction well, and I need some clarification.
Here's the detail. Let $B\mathrm{Top}(n)\subset \mathcal{M}\mathrm{fld}_n$ for the full subcategory spanned by $\mathbb{R}^n$. (By the Kister-Mazur theorem, $B\mathrm{Top}(n)$ is a Kan complex which has the homotopy type of the classifying space of the simplicial group $\operatorname{Sing}\mathrm{Top}(n)=\operatorname{Sing}\operatorname{Homeo}(\mathbb{R}^n,\mathbb{R}^n),$ hence the notation.) The tangent classfier is defined as the composite
$$\mathcal{M}\mathrm{fld}_n\to\mathcal{P}(B\mathrm{Top}(n))\simeq\mathcal{S}_{/B\mathrm{Top}(n)},$$
where the first map is the restricted Yoneda embedding and the second is the straightening-unstraightening equivalence. It is claimed in Corollary 2.13 of Ayala-Francis's article cited above that for any $n$-manifold $M$, $\tau(M)$ may be regarded as the classifying map of the tangent microbundle of $M$. I do not follow their argument very well (due to my incompetence), but roughly, it can be summarized as follows:

*

*Observe that the domain of $\tau(M)$ has the homotopy type of $\operatorname{Sing}(M)$. (This is proved in Proposition 6.1.6 of Yonatan Harpaz's course note also.)


*Observe that $\tau(M)$ classifies the tangent bundle if $M=\mathbb{R}^n$, and conclude by using that the assignment $M\mapsto \pi_M$ is functorial in $M$ by Step 1.
I do not understand Step 2 above; it is not clear to me how the functoriality in $M$ is applied there. (Besides, $\tau({\mathbb{R}^n})$ of course classifies the tangent microbundle, because its domain is contractible by Step 1.) Can anyone explain why $\tau(M)$ classifies the tangent microbundle? Thanks in advance.
 A: For an $\infty$-category $X$ let me explicitly write $\mathrm{Fun}(X,\mathcal{S})$ for $\mathcal{P}(X)$. The construction of the tangent classifier firsts associates to $M$ the presheaf $\mathrm{Emb}(-,M) \in  \mathrm{Fun}(\mathrm{Top}(n),\mathcal{S})$. The result is literally the topological frame bundle $\mathrm{Emb}(\mathbb{R}^n,M)$ with the standard action of $\mathrm{Top}(n)$. It then passes through the equivalence $\mathrm{Fun}(\mathrm{Top}(n),\mathcal{S}) \simeq \mathcal{S}_{/\mathrm{BTop}(n)}$ to arrive at a map $M \rightarrow \mathrm{BTop}(n)$.
This equivalence is not mysterious, at least on the $\pi_0$ level. Taking the product $\mathrm{Top}(n)$, $- \times E \mathrm{Top}(n)$ and equipping it with the diagonal action shows that every $\mathrm{Top}(n)$ space, i.e. an element of $\mathrm{Fun}(\mathrm{Top}(n),\mathcal{S})$, is equivalent to a prinicipal $\mathrm{Top}(n)$ bundle in this category. Up to homotopy, this equivalence of $\infty$-categories is just the classical equivalence between isomorphism classes of principal $\mathrm{Top}(n)$-bundles and homotopy classes of maps into $\mathrm{Top}(n)$. But of course, the principal bundle associated to the topological frame bundle is the tangent microbundle as the tangent microbundle at a point $x$ is essentially given by the germ of the embeddings of $\mathbb{R}^n$ which send the origin to $x$.
