Topological embeddings of real projective space in euclidean space

I was wondering whether the real projective space $\Bbb{R}P^n$ embeds topologically into $\Bbb{R}^{n+1}$ for odd $n$.

It certainly doesn't for even $n$ because of Alexander duality. Also it doesn't embed smoothly for any $n$. I will prove the two above statements in my algebraic topological class, but I couldn't find anything in the literature for topological embeddings in the odd case.

• Just to clarify: you're not assuming any nice behavior, eg locally flat, for your embedding. Right? – Danny Ruberman Jan 28 '18 at 17:50
• If you restrict to locally flat embeddings, it looks like you can rule out $\mathbb{RP}^{4k-1} \hookrightarrow \mathbb{R}^{4k}$. – Marco Golla Jan 28 '18 at 18:48
• The arguments that obstruct a smooth embedding will also obstruct the existence of a locally flat embedding. Put differently, by topological embedding I mean an injective continuous map (that is a homeomorphism onto its image). – Stefan Friedl Jan 28 '18 at 19:51
• – Moishe Kohan Jan 29 '18 at 0:15
• It is essentially a duplicate of an earlier question which was about smooth embeddings but the proof did not use smoothness. – Moishe Kohan Jan 29 '18 at 0:15

The first page of W.~Massey's paper On the imbeddability of the real projective spaces in Euclidean space states that $\mathbb RP^n$ with $n>1$ cannot be imbedded topologically in $\mathbb R^{n+1}$ because its mod $2$ cohomology algebra does not satisfy a certain condition given by R. Thom.
EDIT: Actually, a proof can be found in Is it true that all real projective space $RP^n$ can not be smoothly embedded in $R^{n+1}$ for n >1.
• @StefanFriedl: Mayer-Vietoris sequence works for excisive triads. The solution is to use a fancier cohomology theory, e.g. if $A$, $B$, are closed subsets of the compact Hausdorff space $X$, then the triad $(X,A,B)$ is excisive for the Alexander-Spanier cohomology, see Theorem 8.14, Chapter II, section 8 in a book by W.Massey [Homology and cohomology theory, 1978]. In our case $X=S^{2n}$ and $A$, $B$ are closures of the components of $S^{2n}\setminus\mathbb RP^{2n-1}$. The linked argument is a formal consequence of the Mayer-Vietoris so it applies. – Igor Belegradek Jan 29 '18 at 1:58
• The argument will also need the Alexander duality for the Alexander-Spanier cohomology $H^*$, which can be found in the same Massey's book. Thus $H^{2n}(A)=0=H^{2n}(B)$ with mod $2$ coefficients. – Igor Belegradek Jan 29 '18 at 2:24