TITLE: Topology of a cone of $\mathbb R\mathbb P^2$.
QUESTION [10 upvotes]: I had already posted this on mathoverflow and was advised to post the same here. So here it goes:
$X=\{(x,y,z)|x^2+y^2+z^2\le 1$ and $z≥0\}$ i.e. $X$ is the top half of a $3$-Disk.
$Z=X/E$, where $E$ is the equivalence relation on the the plane $z = 0$ which is as follows:
$(x,y,0)∼(−x,−y,0)$.
I was told that this space is equivalent to a cone of $\mathbb R\mathbb P^2$ (Real Projective Plane).
I want to know the following facts about "$Z$"
1) Is this a manifold with a boundary?
2) If it is a manifold with a boundary, what are the points of $Z$ that make the boundary?
3) Is it simply connected?
4) What is the minimum Euclidean dimension in which $Z$ can be embedded in?
Thank you very much for your help. I am new to topology and this problem came up as a part of my project. Any help is appreciated.
Thank you. Will.

REPLY [2 votes]: This should be a comment somehwere, but got to long.
Will, the cone on $P^2_\mathbb R$ is not a manifold. Consider the long exact sequence for integral reduced homology of the pair $(C,C')$ where $C=C(P^2_\mathbb R)$ is the cone over $P^2_\mathbb R$ and $C'=C\setminus\{a\}$ is the complement of the apex of the cone. Since $C$ is contractible and $C'$ deformation-retracts onto $P^2_\mathbb R$, you get isomorphisms $H^\sharp_2(C,C')\cong H^\sharp_{1}(P^2_\mathbb R)\cong\mathbb Z/2\mathbb Z$. 
If follows that $C$ is not a manifold: in a manifold $M$, for every point $p\in M$ we have that the integral reduced homology $H^\#_\bullet(M,M\setminus\{p\})$ is that of a sphere.
(Generalizing this reasoning, you get a rather strong condition on a manifold for its cone to be also a manifold. I'm sure the topologists among our fellow M.SEers know of a precise characterization.)
In the same way, we see that $C$ is not a manifold with boundary, because in such a space $H^\#_\bullet(M,M\setminus\{p\})$ is, for every $p$, either identically zero or that of a sphere.