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

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.

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.

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.