TITLE: Is every compact $n$-manifold that immerses in $\mathbb{R}^{n+1}$ smoothable?
QUESTION [12 upvotes]: Suppose that $M$ is a compact topological $n$-manifold, and $f: M \rightarrow \mathbb{R}^{n+1}$ is a topological immersion, i.e. a local topological embedding.  Then is $M$ smoothable?  By that, I just mean 'does $M$ carry a compatible smooth structure as a manifold', the particular immersion doesn't have to be 'smoothable' here.
I feel like the answer is yes, but the problem is that if the original immersion doesn't behave well globally then you could get some problems trying to 'smooth its image' using tubular neighborhoods or something.  Maybe a partition of unity argument is enough, though?  I'm not sure how to deal with this problem in higher dimensions, or even in dimension $4$ which is the case I'm the most curious about.
This is related to a followup question in this thread:
https://mathoverflow.net/questions/390922/is-there-a-4-manifold-which-immerses-in-mathbbr6-but-doesnt-embed-in

REPLY [5 votes]: To complement Moishe Kohan's answer, if M is orientable and has dimension $>4$, if the immersion is locally flat (a technical condition required to induce a map on topological tangent bundles), then $M$ is smoothable.
This is an application of some big theorems of differential topology. Let's do the easy part first, if we have an immersion that behaves nice enough to induce a map on topological tangent bundles (sometimes called microbundles), then we can decompose the trivial rank (n+1)-microbundle on $M$ as the sum of the microbundle of $M$ and a rank 1 bundle. From the orientability of $M$, this must be a trivial rank 1 bundle.
Hence, stably the tangent microbundle of $M$ is trivial. One consequence of smoothing theory is that above dimension 4, smooth structures up to concordance are in bijection with lifts of the the stable microbundle to a stable vector bundle structure (up to some equivalence). Since the microbundle is stably trivial, it always has a lift (stably) to the trivial vector bundle. Hence, $M$ is smoothable. I will just note here that this argument implies all (topologically) stably parallelizable manifolds above dimension 4 are smoothable. For some reason I have not been able to find a statement of this kind, but it is an easy corollary of this stable statement of smoothing theory.
I am not sure if codimension 1 immersion always implies there exists a locally flat immersion, I think it is known that there is not necessarily one arbitrarily close, but these things can get very nasty so I wouldn't bet one way or the other.