It is known that any closed topological manifold is homotopy equivalent to an open smooth manifold.

** Question 1.*** What can be said about the smallest
dimension of a smooth manifold
that is homotopy equivalent to a given closed topological manifold?*

The following somewhat heavy-handed argument yields a smooth manifold of roughly twice the dimension, namely, use

- West's solution of Borsuk's conjecture that any compact ANR of dimension $n$, and in particular any closed $n$-manifold, is homotopy equivalent to a finite polyhedron of dimension $\max\{n, 3\}$;
- Stallings-Dranishnikov-Repovs's embedding up to homotopy type theorem that any finite $n$-dimensional polyhedron is homotopy equivalent to a finite $n$-dimensional subpolyhedron of $\mathbb R^{2n}$, so its regular neighborhood is the desired open smooth manifold.

Here is a specific question that shows the state of my ignorance on this matter:

** Question 2. ** * Is there
a closed $n$-manifold which is not homotopy equivalent
to a smooth $(n+1)$-manifold? *

The naive idea to look at the product of a non-smoothable manifold of dimension $\ge 5$ with $\mathbb R$ fails, because such a product is also non-smoothable (by the topological product structure theorem of Kirby-Siebenmann).

**Edit:** Misha kindly corrects me that a $5$-manifold is smoothable if and only if its Kirby-Siebenmann invariant vanishes; in particular, this apples to products of a $\mathbb R$ and a closed $4$-manifold $M$. Thus $M\times\mathbb R$ is smoothable iff
$M$ has zero KS invariant. Smooth $4$-manifolds have zero KS invariant, but amazingly so do some non-smoothable ones.