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Edit:Sorry if this is simple but I have not done homology in a while. We have for all $\mathbb R ^n $and k>0, that $H_k(\mathbb R ^n)$=0 ( trivial group). What would happen if we were to embed in $\mathbb R^4$ a projective 2-space, whose homology is not trivial? Wouldnt this be a cycle that does not bound, contradicting that the host space's homologu is trivial?

MSIS
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  • If you're considering $\mathbb{R}^4 \subseteq \mathbb{RP}^n$ then are you asking to compute $H_k(\mathbb{RP}^n, \mathbb{R}^4)$? – Perturbative May 02 '20 at 21:54
  • Hi, sorry for confusion, just edited. No, inclusion goes the opposite way; we have a projective 2-space embedded in R^4. It seems like an example of a cycle that does not bound, i.e., a non-trivial class, contradicting that Homology of R^4 is trivial. – MSIS May 02 '20 at 22:03
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    Of course it bounds in $\Bbb R^4$. Just take a cone on the surface. A sphere or torus, whose homology is likewise nontrivial, obviously bounds, too. I can’t imagine what you’re thinking of. – Ted Shifrin Sep 29 '21 at 21:27
  • Thank you. Apologies, haven't done any Real Math in a while, obviously :). – MSIS Sep 29 '21 at 21:40

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