Visualization of the quotient of $\mathbb{R}^2$ by an involution. Consider $\mathbb{R}^2$ and let $\mathbb{Z}_2$ act by taking $(x,y) \rightarrow (-x,-y)$ and consider $\mathbb{R}^2/\mathbb{Z}_2$. I can, using algebraic machinery, show that the quotient is the same as the quadric cone in $\mathbb{R}^3$ given by $xy=z^2$ and this is of course all nice, but I am having some trouble visualizing why this is true.
So, my strategy for trying to visualize this gives something which is, unfortunately wrong. I consider circles of radius $r$ and try to see what they're identified as. However, this visualization gives the wrong picture in the end. 
So, are there any nice ways you can think of to visualize why the quotient is the quadric cone?
 A: Taking the real points of a variety does not commute with taking quotients. This is easier to see when you pass to complex points: taking quotients can sometimes identify complex conjugate points, and when that happens they get sent to a new real point in the quotient that isn't visible when you only take the topological quotient of the real points. 
In this particular case, the fixed points of $R = \mathbb{R}[x, y]$ under the given $\mathbb{Z}_2$ action is $R^{\mathbb{Z}_2} = \mathbb{R}[x^2, xy, y^2]$. The points $R^{\mathbb{Z}_2} \to \mathbb{R}$ that are invisible in Donna's answer are points that correspond to sending $x^2$ or $y^2$ to a negative number, and these come from complex rather than real points of $\text{Spec } R$. In fact there is a second invisible plane of complex points in $\text{Spec } R$ both of whose coefficients are purely imaginary. These are precisely the non-real points which get identified with their complex conjugates under the action, and after that identification they are precisely the second half of the cone. 
As subspaces of $\mathbb{C}^2$ the purely real and purely imaginary planes meet at a single point, namely the origin,  so just run Donna's visualization on both of them simultaneously. 
A: You can take the plane, remove an axis, say the $X$ axis. You get two semi-planes. 
Glue the semi-planes according to that relation. This is consists in putting one above the other after rotating half a turn. We get one semi-plane as a result.
Now we take the removed axis and remove the origin. We get two rays. Glue the rays into one ray according to the relation. Glue the resulting ray to the semi-plane where it was supposed to be. We get a cylinder. 
Finally we add the origin, which will close one of the ends of the cylinder.
By the way, I am getting only half of the cone. Are we supposed to consider the whole cone?
A: I can't comment, but the reason for me being suspicous of @Donna's answer is that:  
Consider the action of $\mathbb{Z}_2$ on $\mathbb{R}^2$ as above, and take $Spec \mathbb{R}[x,y]$ with the Zariski topology. This is not the same topology as the analytic one on $\mathbb{R}$, of course. If we take the quotient of $
\mathbb{A}^2_{\mathbb{R}}= Spec \mathbb{R}[x,y]$ by the action taking $x\rightarrow -x$ and $y \rightarrow -y$, then the resulting variety one gets is $\mathbb{R}[x,y,z]/(xy-z^2)$ which is precisely the cone. Now, the real points of the quotient variety will be the orbits of the real points of $\mathbb{A}^2_{\mathbb{R}}$ under the action. This shows that I would expect the answer to be a double cone, not just a cone. 
