TITLE: Is there a $4$-manifold which Immerses in $\mathbb{R}^6$ but doesn't Embed in $\mathbb{R}^7$?
QUESTION [24 upvotes]: I'm interested in both version of the question in the title, i.e. in the topological category and in the smooth category.  By a topological immersion I mean a local embedding.  I was asking in relation to this question:
https://math.stackexchange.com/questions/1801318/dimensions-of-immersions-vs-embeddings
In the very nice answer given in that thread, they work out almost all of the low-dimensional cases for smooth, compact manifolds and smooth immersions/embeddings.  The only 'smooth, compact' case not covered by their answer is the one in the title of this question, so it would be interesting to know if there is a compact example, separately.
As a side-question, for topological $4$-manifolds the case of immersion in $\mathbb{R}^5$ is what remains to be fleshed out in the answer to that previous thread.  Some relevant questions for that case:

Is every compact $4$-manifold which immerses in $\mathbb{R}^5$
smoothable?
Is there a $4$-manifold that immerses in $\mathbb{R}^5$
but doesn't embed in $\mathbb{R}^6$ (resp. $\mathbb{R}^7$)?

By results of Quinn, every open $4$-manifold is smoothable so it's sufficient to prove the smooth case for non-compact manifolds.
I cross-listed one of these on MSE expecting a quick counterexample, but still no takers.  Someone gave a partial answer to the effect of "in high dimensions, codimension-$1$ locally flat immersion implies smoothability":
https://math.stackexchange.com/questions/4115222/is-every-compact-n-manifold-that-immerses-in-mathbbrn1-smoothable

REPLY [21 votes]: Edit: The answer below is incorrect. In fact, $\bar{w}_3(\mathbb{R}P^2\times\mathbb{R}P^2)=0$ (thanks to Rafal Walczak for pointing this out) so by the cited result $\mathbb{R}P^2\times\mathbb{R}P^2$ does embed in $\mathbb{R}^7$! So please strip me of my points and consider the question unanswered...
Edit 2: In fact what the below shows is that the answer to the question in the title is no in the smooth compact case. If $M^4$ immerses in $\mathbb{R}^6$ then $\bar{w}_3(M)=0$, which by the quoted result of Fang implies that $M$ embeds in $\mathbb{R}^7$.

The manifold $\mathbb{R}P^2\times \mathbb{R}P^2$ smoothly immerses in $\mathbb{R}^6$, as a product of Boy's surfaces. However, the main result of
Fang, Fuquan, Embedding four manifolds in (\mathbb{R}^ 7), Topology 33, No. 3, 447-454 (1994). ZBL0824.57014
asserts that a closed smooth $4$-manifold $M$ embeds in $\mathbb{R}^7$ if and only if the normal Stiefel-Whitney class $\bar{w}_3(M)$ vanishes. A quick calculation shows that this is not the case for $M=\mathbb{R}P^2\times \mathbb{R}P^2$, which therefore doesn't embed in $\mathbb{R}^7$.
The cited article also gives necessary and sufficient conditions for topological embeddability of $4$-manifolds in $\mathbb{R}^7$. For example, a closed smooth non-orientable $4$-manifold $M$ admits a locally flat embedding in $\mathbb{R}^7$ if and only if $\bar{w}_3(M)=0$ and $KS(M)=0$, where $KS$ is the Kirby-Siebenmann invariant.