Is real projective space a manifold with boundary? So real projective space,
$$\mathbb{P}^n(\mathbb{R}) \simeq \mathbb{S}^n/\pm,$$
and the upper half-sphere can also be visualized (via bijection) as the upper hemisphere with half of the boundary circle (i.e. $x_i\geq0$). Is the upper hemisphere with half of the boundary circle a manifold with boundary?
This is briefly described in this stackexchange question: Does the real projective space $\mathbb{R}P^n, n \geq 1$ have a boundary?. However, the answers there say that it isn't a manifold with boundary which makes me really confused. Any help would be greatly appreciated!
 A: Remember that a manifold with boundary is a topological space $M$ which is locally homeomorphic to either $\mathbb R^n$ or the half-space $\mathbb H^n:=\{(x_1,\dots,x_n)\in\mathbb R^n:x_1\ge0\}$. Then, one can let the boundary $\partial M$ of $M$ to be the set of points $p\in M$ such that there are no neighborhoods $U\ni p$ homeomorphic to $\mathbb R^n$. Thus, being a manifold (or, equivalently $\partial M=\emptyset$) can be seen as a property of a manifold with boundary $M$.
For any point $[p]\in \mathbb{RP}^n=\mathbb S^n/\sim$, the neighborhood $U=\{[q]\in\mathbb{RP}^n:p\cdot q\ne 0\}$ is homeomorphic to $\mathbb R^n$, where $U$ is well-defined since $p\cdot q\ne0$ iff $p\cdot(-q)\ne0$ for $p,q\in \mathbb S^n$. Indeed, choosing $p_1,\dots,p_n\in\mathbb S^n$ such that $p,p_1,\dots,p_n$ are mutually orthogonal, then there is a homomorphism to the open ball in $\mathbb R^n$
$$U\to B_1(0):[q]\mapsto(q\cdot p_1,\dots,q\cdot p_n),$$
where the representative $q\in[q]$ is chosen (uniquely) so that $q\cdot p>0$. Thus, $\mathbb{RP}^n$ is a manifold, i.e., has no boundary.
