Is the inclusion of the "smooth singular Kan complex" a homotopy equivalence?

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Let $X$ be a smooth manifold.

Let $L$ be the singular simplicial set of $X$. We know $L$ is a Kan complex.

Let $K \subset L$ be the "simplicial subset" consisting of only those singular simplices (in $X$) which are smooth (as a map from a topological simplex to $X$). (Here we view the topological $n$-simplex as a smooth manifold with corners.) We note $K$ is also a Kan complex.

My question is whether the inclusion $K \hookrightarrow L$ is a homotopy equivalence (of simplicial sets / Kan complexes)? How would this be demonstrated?

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Yes. This follows from the Whitney approximation theorem, which says any continuous map between smooth manifolds is homotopic to a smooth map, which can be chosen to coincide with the original map on a closed subset where it is already smooth. (The latter part implies then that two smooth maps are continuously homotopic iff they are smoothly homotopic.) So, homotopy classes of maps $S^n\to X$ are in bijection with smooth homotopy classes of smooth maps $S^n\to X$ via the natural inclusion. The former are the homotopy groups of $L$ and the latter are the homotopy groups of $K$, so this says the inclusion $K\to L$ is a weak equivalence.