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).