Is every compact $n$-manifold that immerses in $\mathbb{R}^{n+1}$ smoothable? 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
 A: Here is a possible strategy on how to find such an example (non-smoothable, immersible).

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*Find a closed 4-manifold $M$ with $k(M)=0$ (here $k$ is the Kirby-Siebenmann invariant), but non-smoothable (by Donaldson's theory or via SW invariants), $\pi$-manifold (see below).


*Recall that $k(M)=0$ implies that $M\times {\mathbb R}$ is smoothable. For instance, if $M$ is the connected sum of two $E8$-manifolds, then $M$ is not smoothable (Donaldson), but $k(M)=0$ (since $k$ is additive under the connected sum).
Definition. A manifold $M$ is called a $\pi$-manifold (I think, the terminology is due to Hirsch) if $M\times {\mathbb R}$ is smooth and parallelizable.
Now, by the Smale-Hirsch theorey, each parallelizable manifold admits an immersion in the equidimensional Euclidean space. Thus, if you manage to find $M$ satisfying all these conditions, there is an immersion $M\times {\mathbb R}\to {\mathbb R}^5$, hence, a flat immersion $M\to {\mathbb R}^5$. At the same time, $M$ is not smoothable.
Unfortunately, $M$ equal to the sum of two $E8$-manifolds is not a $\pi$-manifold, so one needs to dig deeper. What you need is $M$ which is spin and has zero signature  then $M$ should be  be a $\pi$-manifold. (Since spin implies almost parallelizable, while vanishing of the signature implies that $p_1(M)=0$,  together these should imply a $\pi$-manifold, see Proposition 8.4 of A.Kosinski, "Differential Manifolds". The caveat is that the proof that we get a $\pi$-manifold assumes that $M$ is smooth; it remains to be checked check if the proof can be modified to work with topological tangent bundle.)
A: 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.
