Taking a quotient of the 1-sphere by identifying diametrically opposite points

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?

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You are identifying every point of the circle with its antipode, so you are not getting anything looking like a figure 8! – Mariano Suárez-Alvarez Jan 3 '13 at 4:19
OTOH, I don't understand what you mean by the paragraph starting with «It's not hard to see...» but surely it does not prove what you seem to think it does! – Mariano Suárez-Alvarez Jan 3 '13 at 4:29
The relation $\,x\sim -x\,$ is not reflexive so it can hardly be an equivalence one... – DonAntonio Jan 3 '13 at 4:44
@Mariano: Regarding your first comment: I'm aware of that, as shown in the sentence following the one about the figure-eight. Regarding the second: this question was the result of paring down a larger one, and in the process of cutting things out things got confused. Thanks for the help down below. – Alex Petzke Jan 3 '13 at 21:28
@DonAntonio: The book takes the convention that a point is always considered equivalent to itself. – Alex Petzke Jan 3 '13 at 21:30

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.

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Let $\sim$ be the equivalence relation. I will identify $S^1$ with the set of complex numbers of modulus $1$.

• Consider the function $f:z\in S^1\mapsto z^2\in S^1$.
• It is clear that if $x$, $y\in S^1$ are such that $x\sim y$ then $f(x)=f(y)$, for in that case we have $x=\pm y$. This has the consequence that there is a function $\bar f:S^1/\mathord\sim\to S^1$ such that $\bar f([z])=z^2$ for all $z\in s^1$. Properties of the quotient topology imply at once that $\bar f$ is a continuous function. Check all this in detail!
• One can easily see that if $x$, $y\in S^1$ are such that $f(x)=f(y)$ then $x=\pm y$. This has as a consequence the fact that $\bar f$ is injective. Check this in detail!
• Finally, $\bar f$ is a surjective function —this is a consequence of the fact that $f$ itself is surjective.
• At this point, we got outselves a continuous bijection $\bar f:S^1/\mathord\sim\to S^1$.
• Now, there is a theorem which tells us that

a continuous bijection from a compact space to a Hausdorff space is an homeomorphism.

• Using this, we get that $\bar f$ is an homeo.
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Start with a circle and fold it into a figure eight as shown above. Now fold it along the vertical centreline so that $A$ and $A'$ coincide, as do $B$ and $B'$, and $C$ and $C'$. You’ve now identified each point on the original circle with the point that was diametrically opposite it, and the result is a single circle.

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This is the most visually appealing argument that I've seen so far. – Haskell Curry Jan 3 '13 at 6:40

The way I think of it is, if you've identified every point with its antipode, then your original circle now has two copies of each point you care about. So let's throw away the duplicates and just keep half the points, giving you a semicircle. Except, oops, the two ends of the semicircle are also the same. So glue them together and you get a circle again.

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