I do not have any clue to the following question: What is the homology group of $(\mathbb{R}^n \times \mathbb{R}^n \setminus \Delta) / (\mathbb{Z} / 2)$, where $\Delta$ is the diagonal in $\mathbb{R}^n \times \mathbb{R}^n$, and the equivalence relation is $(x, y) \sim (y, x)$ for all $x \ne y$.
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What kind of homology are you taking? – AnonymousCoward Dec 05 '17 at 19:27
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It's not hard to see that the space in question is homotopy equivalent to something familiar. – AnonymousCoward Dec 05 '17 at 19:28
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@AnonymousCoward thanks! I got to the step where $\mathbb{R}^n \times \mathbb{R}^n \setminus \Delta \sim S^{n - 1}$ but couldn't get around to get the (singular) homology of $S^{n - 1} / (\mathbb{Z} / 2)$. – Riley Zhang Dec 05 '17 at 19:33
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- There is an obvious $Z/2$ invariant deformation retract of $R^n\times R^n/\{(x,x)\}$ onto the orthogonal complement of the diagonal subspace, $V = \{(x,-x)\}$, and there is a further $Z/2$ invariant deformation retract in this subspace of $V\setminus \{0\}$ onto the unit sphere in $V$.
- Hint: What is the described action of $Z/2$ when restricted to $V$? It coincides with a familiar action when you identify $V$ with $\mathbb{R}^n$ in the "obvious" way.
- Further hint: What is the most familiar $Z/2$ quotient of a sphere you can think of?
AnonymousCoward
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For the final hint, is the $\mathbb{Z} / 2$ quotient of $S^{n - 1}$ $\mathbb{RP}^{n - 1}$? – Riley Zhang Dec 05 '17 at 21:12
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What is the action of $Z/2$ on $S^{n-1}$ that yields $RP^{n-1}$ as the quotient? Can you show that the $Z/2$ action you have here is the same as that one? If so, then your answer is yes. – AnonymousCoward Dec 05 '17 at 21:15