I have been working on the following problem:
"Let $\sim$ be the equivalence relation on the unit circle $S^1$ defined by $x \sim -x$, $x \in S^1$. Show that $S^1/\sim$ is homeomorphic to $S^1$ and interpret geometrically."
I have applied the following two theorems:
"Let $X$ and $Y$ be spaces and $f:X \to Y$ a continuous function from $X$ onto $Y$. In order that the natural correspondence $h:X/\sim_f \to Y$ defined by $h([x])=f(x)$, $x \in X$ be a homeomorphism, it is necessary and sufficient that $Y$ have the quotient topology determined by $f$."
"Let $X$ and $Y$ be spaces and $f:X \to Y$ a continuous function from $X$ onto $Y$. If $f$ is either open or closed, then $Y$ has the quotient topology determined by $f$."
It's not hard to see that $f(x)=-x$ is continuous, surjective, and open, so that $S^1/\sim\,\,\cong S^1$ follows from there.
I'm having a difficult time picturing this geometrically. If we identify two opposite points on the circle, the resulting space would resemble a pinched-together circle, or figure-eight. I can't picture how pinching every pair of opposite points together yields the circle again. Could someone explain the geometric intuition behind this?
Best Answer
Imagine your circle lying in the $xy$-plane in $3$-space. Now pinch the points $(0,\pm1)$ together to the origin, giving, as you say, a figure-eight. Now take only the right-hand loop and, in space, rotate it $180$ degrees in the $x$-axis. Now take this loop and flip it, in space, through the $y$-axis; that is, fold it over so that all the picture is in the left-hand half-plane. If you follow what happened to any point originally at $(x,y)$ with $x>0$, you see that it lands on the point whose original coordinates were $(-x,-y)$. Voilà.