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The Join of Two Circles Is the Three Sphere

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
823
India
Problem. I want to prove that $S^1*S^1$ is homeomorphic to $S^3$, that is, the join of two copies of $S^1$ is homeomorphic to $S^3$.

(Writing $I$ to denote the closed unit interval, the join of two spaces $X$ and $Y$ is defined as $(X\times Y\times I)/\sim$, where $\sim$ is an equivalence relation which identifies $(x, y_1 0)$ with $(x, y_2, 0)$ and $(x_1, y, 1)$ with $(x_2, y, 1)$ for all $x_1, x_2\in X$ and $y_1, y_2\in Y$.)

I tired the following. Think of $S^1$ as $I/\partial I$, and write $\pi:I\to S^1$ to denote the natural projection map. Then we have a map $f: (I\times I)\times I\to (S^1\times S^1)\times I$ which sends $(x, y, t)$ to $(\pi(x), \pi(y), t)$. Let $q: (S^1\times S^1)\times I\to S^1*S^1$ be the natural map coming from the equivalence relation $\sim$.

Thus we have a surjective continuous map $q\circ f: (I \times I)\times I\to S^1*S^1$. Write $q\circ f$ as $g$. Since the domain of $g$ is compact, we know that if $\simeq_g$ is the equivalence relation on $I^3$ induced by $g$, then $I^3/\simeq_g$ is homeomorphic to $S^1*S^1$.

I was sure that $\simeq_g$ would turn out to be such that it identifies all points of $\partial I^3$ and no point in the "interior" of $I^3$ with any other point. So that we would have $I^3/\simeq_g = I^3/\partial I^3$, which is homeomorphic to $S^3$.

But to my surprise this is not the case! For example, consider the points $p:=(1/2, 1/2, 0)$ and $q:=(1/2, 1/2, 1)$ in $I^3$. Then $g(p)\neq g(q)$.

What is weirder is that $\simeq_g$ does make $I^3/\simeq_g$ homeomorphic to $S^3$ nevertheless, despite the fact that equivalence classes induced on $I^3$ by $\simeq_g$ are finer that the equivalence classes on $I^3$ induced by identifying all points in $\partial I^3$ to one point.

Can anybody please provide a proof and if possible comment on the (apparent) weird phenomenon happening above (or point out a mistake somewhere).

Thank you.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,643
Leiden
Hey caffeinemachine !

Isn't the first a torus and the second a sphere?
What's the reason you think they may be homeomorphic?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
823
India
Hey caffeinemachine !

Isn't the first a torus and the second a sphere?
What's the reason you think they may be homeomorphic?
Hey.

By $S^1*S^1$ I do not mean $S^1\times S^1$. $S^1*S^1$ is the "join" of two copies of $S^1$ (which I have defined in the post).

This is an exercise in Hatcher. More generally Hatcher has asked to prove that $S^m*S^n$ is homeomorphic to $S^{m+n+1}$.
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Hi caffeinemachine ,

Here is an outline. Consider $\Bbb S^3$ as the set of all ordered pairs $(x,y)\in \Bbb R^2 \times \Bbb R^2$ such that $\|x\|^2 + \|y\|^2 = 1$. Define a map $S^1 \times S^1\times I \to S^3$ by sending a point $(a,b,t)$ to $\left(a\sin \dfrac{\pi t}{2}, b\sin \dfrac{\pi(1-t)}{2}\right)$. This is a continuous map that respects the equivalence relation ~, so there is a continuous map $\Bbb S^1 * \Bbb S^1\to \Bbb S^3$ such that $[(a,b,t)] \mapsto \left(a\sin \dfrac{\pi t}{2}, b\sin \dfrac{\pi(1-t)}{2}\right)$. Its inverse is the map $\Bbb S^3 \to \Bbb S^1 * \Bbb S^1$ sending $(x,y)$ to $\left[\left(\dfrac{x}{\|x\|}, \dfrac{y}{\|y\|}, \dfrac{2}{\pi}\sin^{-1}\lvert x\rvert\right)\right]$. Hence, $S^1 * S^1 \to \Bbb S^3$ is a bijective continuous map; the spaces involved are compact Hausdorff spaces, so this map is a homeomorphism from $\Bbb S^1 * \Bbb S^1$ to $\Bbb S^3$.


The same argument shows $\Bbb S^m * \Bbb S^n$ is homeomorphic to $\Bbb S^{m+n+1}$.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
823
India
Hi caffeinemachine ,

Here is an outline. Consider $\Bbb S^3$ as the set of all ordered pairs $(x,y)\in \Bbb R^2 \times \Bbb R^2$ such that $\|x\|^2 + \|y\|^2 = 1$. Define a map $S^1 \times S^1\times I \to S^3$ by sending a point $(a,b,t)$ to $\left(a\sin \dfrac{\pi t}{2}, b\sin \dfrac{\pi(1-t)}{2}\right)$. This is a continuous map that respects the equivalence relation ~, so there is a continuous map $\Bbb S^1 * \Bbb S^1\to \Bbb S^3$ such that $[(a,b,t)] \mapsto \left(a\sin \dfrac{\pi t}{2}, b\sin \dfrac{\pi(1-t)}{2}\right)$. Its inverse is the map $\Bbb S^3 \to \Bbb S^1 * \Bbb S^1$ sending $(x,y)$ to $\left[\left(\dfrac{x}{\|x\|}, \dfrac{y}{\|y\|}, \dfrac{2}{\pi}\sin^{-1}\lvert x\rvert\right)\right]$. Hence, $S^1 * S^1 \to \Bbb S^3$ is a bijective continuous map; the spaces involved are compact Hausdorff spaces, so this map is a homeomorphism from $\Bbb S^1 * \Bbb S^1$ to $\Bbb S^3$.


The same argument shows $\Bbb S^m * \Bbb S^n$ is homeomorphic to $\Bbb S^{m+n+1}$.
This is a nice argument. Thanks!
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
West Bengal, India
Think of the two $S^1$'s as $[0, 1]/0 \sim 1$. $S^1 * S^1$ is then literally $[0, 1] \times [0, 1] \times [0, 1]$ with $[0, 1] \times [0, 1] \times \{0\}$ collapsed to the first $[0, 1]$ and $[0, 1] \times [0, 1] \times \{1\}$ collapsed to the second $[0, 1]$, and there's a further identification between (1) $\{0\} \times [0, 1] \times [0, 1]$ and $\{1\} \times [0, 1] \times [0, 1]$ and (2) $[0, 1] \times \{0\} \times [0, 1]$ and $[0, 1] \times \{1\} \times [0, 1]$.

Here's a picture of $[0, 1] * [0, 1]$ :

220px-Join.svg.png

$S^1 * S^1$ is then obtained from identifying pairs of opposite sides of the tetrahedron (the ones sharing either the red or the green edge). To see why this is $S^3$, cut along the little square in the middle to get two "pieces of pie", and identify the opposite sides in each "piece of pie" accordingly. These become solid torii, with boundary being the torus obtained by identifying the little squares appropriately. Thus $S^1 * S^1$ is just union of two solid torii glued along the boundary torii by a homeomorphism $T^2 \to T^2$ taking meridians to longitudes and vice versa.

So it remains to prove $S^3$ can be decomposed this way. But this is not hard; $S^3 = \partial(D^4) = \partial(D^2 \times D^2) = S^1 \times D^2 \cup_{S^1 \cup S^1} D^2 \times S^1$. The gluing map along $S^1 \times S^1$ is precisely the one switching the roles of the meridians and the longitudes.