What is the quotient space of ${\mathbb C}^n$ by the antipodal action Consider the ${\mathbb Z}_2$ action on ${\mathbb C}^n$ given by
$[z_0,z_1,\cdots, z_n]\mapsto [-z_0, -z_1,\cdots, -z_n]$.
My question is what is the quotient space? Is it homeomorphic to ${\mathbb C}^n$?
For $\mathbb C$ the quotient space is the upper half-plane with positive $x$-axis is identified with the negative $x$-axis, which is again homeomorphic to $\mathbb C$.
 A: No it's not $\mathbb C^n$ for $n\geq 2$
You're essentially looking at $\mathbb R^{2n}$ with the antipodal action, which is free on $\mathbb R^{2n}\setminus\{0\}$, but $0$ is a fixed point so there'll be something odd going on there. 
Let's call the quotient $X$ and let $ Y= X\setminus \{0\}$. Then $Y$ is the quotient of $\mathbb R^{2n}\setminus \{0\}$ (this is a little exercise) and so it's a manifold with $\pi_1(Y) = \mathbb{Z/2Z}$ if $n\geq 2$ (here we use that for $n\geq 2$, $\mathbb R^{2n}\setminus \{0\}$ is simply connected)
You can now clearly see that $X$ is not $\mathbb C^n$ if $n\geq 2$ : there's no point $x$ of $\mathbb C^n$ such that $\pi_1(\mathbb C^n\setminus \{x\}) \cong \mathbb{Z/2Z}$. 
In fact our earlier computation and description should give us some hint as to what $Y$ is: its homotopy groups are $\pi_1 =\mathbb{Z/2Z}$ and  $\pi_k = \pi_k(\mathbb R^{2n}\setminus\{0\}) = \pi_k(S^{2n})$ for $k\geq 2$, so that $Y$ has the same homotopy groups as $\mathbb RP^{2n-1}$
Now define $S^{2n-1}\to \mathbb R^{2n}\setminus \{0\}$ to simply be the inclusion, then this is of course comaptible with the $\mathbb Z/2$-actions and so induces a map $\mathbb RP^{2n-1}\to Y$; and in fact the usual retraction $\mathbb R^{2n}\setminus \{0\}\to S^{2n-1}$ is $\mathbb{Z/2Z}$-equivariant as well (and so is the homotopy that exhibits it as a strong deformation retraction) and so it yields that $Y$ strongly deformation retracts onto $\mathbb RP^{2n-1}$
So $X$ deformation retracts onto $\mathbb RP^{2n-1}$ when you remove $0$ : it's a weird looking space, I don't know if there's an easy description of it (although I'll be glad to see in other answers if there is)
EDIT : Thinking about it for a bit, I actually found a better description of $X$. Note, again, that $X$ is the quotient of $\mathbb R^{2n}$ by the antipodal action. Now $\mathbb R^{2n}$ is homeomorphic to $(\mathbb R_+\times S^{2n-1})/\{0\}\times S^{2n-1}$ (that's an easy exercise of general topology), and the antipodal action corresponds to the antipodal action on $S^{2n-1}$. 
From there it's easy to check that we get a continuous map $X\to (\mathbb{R_+\times R}P^{2n-1})/\{0\}\times \mathbb RP^{2n-1}$ which is a bijection on sets.  Then it's also easy to check that the converse map is also continuous, so this is a description of $X$ : if we remove the problematic point, we get something that's homeomorphic to $\mathbb R_+^* \times \mathbb RP^{2n-1}$ and so indeed retracts onto $\mathbb RP^{2n-1}$ as claimed above. 
As Jason DeVito points out, this shows that $X$ is the (noncompact) cone on $\mathbb RP^{2n-1}$ : $C\mathbb RP^{2n-1}$
Note that in the $n=1$ case, we recover $\mathbb R_+\times \mathbb RP^1/...$ and $\mathbb RP^1 = S^1$ so we get $\mathbb R_+\times S^1/... \cong \mathbb C$ : we recover that as well. 
(actually if you know some category theory, the proof becomes easier : $\mathbb{R_+}\times S^{2n-1}/(\{0\}\times S^{2n-1})$ is a pushout of $\mathbb Z/2$-spaces ($\mathbb R_+ \times S^{2n-1} \coprod_{\{0\}\times S^{2n-1}} *$) and colimits commute with colimits, therefore the quotient by the $\mathbb Z/2$ action is the pushout of the quotients; clearly the quotient of $\{0\}\times S^{2n-1}$ is $\{0\}\times \mathbb RP^{2n-1}$ and $\mathbb R_+$ is locally compact so $\mathbb R_+\times -$ preserves colimits as well, so that the quotient of $\mathbb R_+\times S^{2n-1}$ is $\mathbb R_+\times \mathbb RP^{2n-1}$. So we reduce the point-set topology to the description of $\mathbb R^{2n}$ as a pushout, instead of having to do point-set topology twice)
