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Consider the standard solid ball $\{(x,y,z)\in \mathbb{R}^3\mid x^2+y^2+z^2\le 1\}$, and the equivalence relation doing nothing in the interior and identifying antipodal points of the boundary. What does the quotient space look like? Is it a manifold? What does its boundary (if it has one) look like?

It seems to me that the equivalence relation can be related to that generating the real projective plane $P^2(\mathbb R)$; in that case the action is free and properly discontinuous, so that the quotient is a manifold. But it this case, it seems I built something whose boundary is $P^2(\mathbb R)$, a well-known contradiction.

Where am I wrong?

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up vote 8 down vote accepted

What you get when you do that quotient is a space homeomorphic to $\mathbf P^3(\mathbb R)$. In particular, it does not have a boundary.

A way to see this is to remember that $\mathbf P^3(\mathbb R)$ is more usually built as the quotient space of a $3$-sphere $S^3\subseteq\mathbb R^4$ by identifying antipodal points, and noticing that when you do this, the closed northern hemisphere $S^3_+\subset S^3$ is a fundamental domain, so that $\mathbf P^3(\mathbb R)$ can also be obtained from $S^3_+$ by quotienting by the induced equivalence relation. Massage things a bit, and you get your construction.

If you go one dimension down, all this should be much easier to visualize: build an intuitive picture there and then go back up one dimension.

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