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.

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 $S^{m+n+1}$.

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 $\Bbb S^{m+n+1}$.

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 $\left(a\sin \dfrac{\pi t}{2}, b\sin \dfrac{\pi(1-t)}{2}\right)$. 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!