Stable smoothing of topological manifolds relative to an embedding Let $M$ be a topological manifold. We know that $M$ is stably smoothable if and only its tangent microbundle, up to stabilization, admits a reduction to vector bundle.
Now I wonder if there is a relative version of this fact. Suppose $M$ and $N$ are topological manifolds and $f: M \to N$ is a locally flat topological embedding and assume it admits a normal microbundle. We also assume that the normal microbundle is equivalent to a vector bundle. Suppose $M$ and $N$ are both stably smoothable. I wonder if one can prove that there are smoothable stablizations $M = M \times {\mathbb R}^m$ and $N' = N \times {\mathbb R}^{m + n}$ such that the natural extension $f': M' \to N'$ (extending $f$ by the inclusion ${\mathbb R}^m \to {\mathbb R}^{m+n}$) is a smooth embedding.
If the above speculation is true, I also wonder if such stable smoothings are in any sense unique.
 A: I think the answer is yes, it follows from this: if $M$ is a triangulable manifold of dimension $n$ greater than 4, and if the total space $E$ of a vector bundle over $M$ is smoothable, then the smooth structure is concordant to one where the zero section is a smooth submanifold.
To see why it follows, stabilize $M$ so that it has dimension at least $5$. Then the normal bundle of the embedding satisfies the requirements of our theorem since the tubular neighborhood is assumed to be a vector bundle and is an open subset of a smoothable space. By the above proposition and the concordance extension theorem, we can find a concordant smooth structure on $N$ which agrees with the smooth structure on the tubular neighborhood of $M$ which we have manufactured to have $M$ as a submanifold.
So why is the first proposition true? Pick a triangulation of $M$. Over an n-simplex $\Delta$ we inductively apply the fact that any smooth structure on $\Delta \times \mathbb{R}^k$ which is a product structure when restricted to some set of faces $I$ is concordant relative $I \times \mathbb{R}^k$ to a product structure. This is a consequence of the product structure theorem. After this process is completed, the smooth structure on our vector bundle has the property that the restriction to every $n$-simplex in $M$ is a smooth submanifold. Hence, $M$ is a smooth submanifold. The vector bundle requirement ensures that any two trivializations over a shared face $\sigma$ induce the same smooth structure on $\sigma \times \mathbb{R}^k$, meaning the notion of a local smooth product structure on a vector bundle is well defined.
