Does identifying opposite points in Euclidean space result in a smooth manifold? Taking Euclidean space $\mathbb{R}^n$ and identifying all pairs of points $\{\mathbf{x}, -\mathbf{x}\}$ results in a topological quotient space $\mathbb{R}^n/\mathbb{Z}_2$. Is this quotient space a smooth (Riemannian, etc.) manifold? If so, does that manifold have a simpler or more standard equivalent description? What is its algebraic topology? Intuitively, it seems to me that the quotient space should be a smooth manifold, except possibly at the origin. (If we'd started off with the sphere $S^n$ instead of Euclidean space $\mathbb{R}^n$, then we would of course get the real projective space smooth manifold $\mathbf{RP}^n$.)
A few further thoughts: if we foliate $\mathbb{R}^n$ into concentric spheres, then we see that this topological space consists of a bunch of nested copies of $\mathbf{RP}^n$. So away from the origin it should locally look like $\mathbf{RP}^n \times \mathbb{R}$ and so should be a smooth manifold, but I'm not sure what happens at the origin. Another way to approach the problem is to distinguish one Cartesian coordinate (WLOG let it be the first one) and think of $\mathbf{x} \in \mathbb{R}^n$ as a direct sum $(x^1, \mathbf{y}) \in \mathbb{R} \oplus \mathbb{R}^{n-1}$. We can then forget about the half-space $x^1 < 0$ since it's identified with the other half-space (with the appropriate inversion of the orthogonal subspace). If we denote the quotient space $\mathbb{R}^n / \mathbb{Z}_2$ by $Y_n$, then the identification reduces the boundary hyperplane $(x^1 = 0) \cong \mathbb{R}^{n-1}$ to a lower-dimensional version of the same problem, so we can recursively describe $Y_n$ as an $n$-dimensional Euclidean half-space bounded by $Y_{n-1}$. I suspect there's a simpler way to think about it though.
P.S. I'm just a dumb physicist without much background in advanced math. I'd appreciate any visual intuition for the $n = 2$ and $n = 3$ cases.
 A: This is not smooth in general. For $n = 2m$ even there is a nice description in term of algebraic geometry : take $\Bbb C^N$ with coordinates $z_1, \dots, z_m, w_{1,2}, w_{1,3}, \dots, w_{m-1,m}$. The corresponding equations are $z_iz_j = w_{ij}^2$. For example, $m=2$ gives the quadratic cone $xy = z^2$, singular.
In general, I claim your space is topologically the cone over $\Bbb RP^{n-1}$, that is the quotient space $[0,1) \times \Bbb RP^{n-1} /\sim$, where $(0,x) \sim (0,y)$ for all $x,y \in \Bbb RP^{n-1}$. 
To see that $\Bbb R^n/(\Bbb Z/2\Bbb Z)$ is the cone over a real projective space, we look at the map $ p : X:= \Bbb R_{\geq 0} \times S^{n-1} \to \Bbb R^n$ contracting $S^{n-1} \times \{0\}$ to the origin. There is a natural $\Bbb Z/2 \Bbb Z := \Gamma$-action on $X$ and $p$ is $\Gamma$-equivariant. This means that we get a map $p : X/\Gamma \to \Bbb R^n/\Gamma$. This map is clearly surjective and proper. This means that we get an isomorphism $p(X/\Gamma) \cong \Bbb R^n/\Gamma$. But $p$ is exactly contracting $S^{n-1}$ to a point, so we recover the previous description. 
The intuition is as follow : $p$ is replacing $0$ by $E := S^{n-1}$, call the new space $X$. Now, taking the quotient of $\Bbb R^n$ or taking the quotient of $X$ and then contracting $E$ changes nothing. But the description with $\Bbb R_{\geq 0} \times S^{n-1}$ makes clear that you are just taking a bunch of $\Bbb RP^{n-1}$ as you said, and adding the origin at the end. 
