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Sep 4 at 12:57 vote accept Thorgott
Sep 4 at 12:57 comment added Thorgott Thanks again, should've seen that. I'll note that taking $M=S^1$, $N=2$ and $U=\mathbb{R}^2\setminus\{0\}$, this method reproduces the example in @tkf's answer.
Sep 4 at 9:08 comment added Connor Malin @Thorgott One point compactification is contravariantly functorial with respect to open embeddings; if $U \rightarrow \mathbb{R}^N$ is an open embedding then the map $(\mathbb{R}^n)^+ \cong S^N \rightarrow U^+$ is given by $x \rightarrow x$ if $x \in U$ and otherwise $\infty$. The map is an isomorphism on $H^N$ since the degree of the map (preimage of a generic point) is 1.
Sep 4 at 0:09 comment added Thorgott Thanks for answering. I understand the $S^{N-n}\rtimes M$ is the one-point compactification of such an open tubular neighborhood, but how in general do we construct the map from $S^N$ that is non-trivial on $H_N$?
Sep 2 at 21:11 history answered Connor Malin CC BY-SA 4.0