The space obtained by identifying the antipodal points of a circle Take the unit circle $S^1$ in $\Bbb{R}^2$ and partition it into subsets which contain exactly two points, the points being antipodal (at opposite ends of a diameter). $P$ is the resulting identification space. 
Now, what is homeomorphic to $P$? My intuition says that it must be a semicircle. But couldn't construct a homeomorphism.
 A: I think your space is homeomorphic to the real projective line $\mathbb{R}P^1$.
$\mathbb{R}P^1$ is space of one dimensional subspaces of $\mathbb{R}^2$. Look at wiki page, as you can see real projective line is homeomorphic to $S^1$. So what you get after your construction is circle again.
In fact if you do your construction in more dimensions. Take $S^n$ and glue antipodal points and you get $\mathbb{R}P^n$. But for example in two dimensions $\mathbb{R}P^2$ is not homeomorphic to sphere since real projective plane has genus 1.
A: Identify antipodal points on the circle:

Form the identification space:

Moving along this space, you come back to where you started... just like on a circle:

(original gifs here and here; I joined them because when they load at different times in the browser, they don't stay in sync)
This correspondence between the points of $P$ and the points of $S^1$ is a homeomorphism.

PlotUpTo[angle_, colorscale_] := 
 ParametricPlot[{Cos[t], Sin[t]}, {t, 0, angle}, PlotStyle -> Thickness[0.05],
 ColorFunctionScaling -> False,
 ColorFunction -> Function[{x, y}, Hue[colorscale (Arg[-x - I y] + Pi)/(2 Pi )]],
 PlotRange -> All, PlotRangeClipping -> False, ImageSize -> {300}, Axes -> None]

AnimatedMarker[time_, speed_] := Graphics[{PointSize[0.05], 
 Point[{Cos[speed 2 Pi time], Sin[speed 2 Pi time]}],
 White, PointSize[0.03], Point[{Cos[speed 2 Pi time], Sin[speed 2 Pi time]}]}]

AnimatedPlot[angle_, colorscale_, time_, speed_] := 
 Show[PlotUpTo[angle, colorscale], AnimatedMarker[time, speed]]

Export["animation.gif", Table[Rasterize[
 TableForm[{{AnimatedPlot[Pi, 2, t, 1/2]}, {AnimatedPlot[2 Pi, 1, t, 1]}},
 TableSpacing -> {10, 0}]], {t, 0, 1, 0.01}], "DisplayDurations" -> {0.04}]

A: Start with $S^1$, as in the first figure below. Hold it at $c'$ and $c$, and flip the righthand side over to form an $\infty$ sign, as in the second figure. Now fold the loops together into a single loop, as in the third figure. Everything matches up properly to give you your quotient space: it’s $S^1$ again.

A: Consider the (complex) function $f(z) = z^2$ from the circle onto itself (seeing the plane as the complex plane). This identifies exactly the antipodal points on the circle. So this map induces a homeomporphism between $S^1$ modulo this equivalence and $S^1$.
