
MHB Journeyman
#1
October 16th, 2016,
04:16
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.

October 16th, 2016 04:16
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MHB Seeker
#2
October 18th, 2016,
22:18
Hey caffeinemachine!
Isn't the first a torus and the second a sphere?
What's the reason you think they may be homeomorphic?

MHB Journeyman
#3
October 18th, 2016,
23:33
Thread Author
Originally Posted by
I like Serena
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 $S^{m+n+1}$.

MHB Master
#4
October 20th, 2016,
13:34
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 $\left(a\sin \dfrac{\pi t}{2}, b\sin \dfrac{\pi(1t)}{2}\right)$. 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 $[(a,b,t)] \mapsto \left(a\sin \dfrac{\pi t}{2}, b\sin \dfrac{\pi(1t)}{2}\right)$. 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 $\Bbb S^{m+n+1}$.

MHB Journeyman
#5
October 20th, 2016,
15:01
Thread Author
Originally Posted by
Euge
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 $\left(a\sin \dfrac{\pi t}{2}, b\sin \dfrac{\pi(1t)}{2}\right)$. 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 $[(a,b,t)] \mapsto \left(a\sin \dfrac{\pi t}{2}, b\sin \dfrac{\pi(1t)}{2}\right)$. 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!

MHB Journeyman
#6
December 24th, 2016,
05:38
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 $[0, 1] * [0, 1]$ :
$S^1 * S^1$ 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.