Join of $S^1$ with $S^1$ gives $S^3$. 
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.
 A: The map $g$ induces the following identifications
$$
(x,y,0) \sim (x,y',0), \quad (x,0,z) \sim (x,1,z), \quad (x,y,1) \sim (x',y,1), \quad (0,y,z) \sim (1,y,z)
$$
for all $x,x',y,y',z \in I$. The second and the fourth relation come from the usual identifications turning the unit square into a torus, while the first and the third relation stem from the construction of the join. The first identification degenerates one side of the cube to a line, so we get a "prism", and the second identification folds the $(y=0)$-side to the $(y=1)$-side, which gives us a solid cylinder. The $(z=1)$-side of the cube has now become the hollow cylinder, and here each interval ranging from a point in one end of the cylinder to its opposite point in the other end is now collapsed to a point. This yields a ball $D^3$. As a last step, we identify every point in one half-sphere of the ball's boundary to the point in the other half-sphere which has the same $y$ and $z$ coordinates, and this produces $S^3$. If you have problem visualizing the last step, go down one dimension and think of the disk $D^2$. When we identify points in $\partial D^2$ that lie above each other we are basically folding the disk around a sphere $S^2$, and the boundary becomes a line from one point in the sphere to its opposite point.
The phenomenon you describe is nothing to worry about. It is well possible that a coarser relation produces the same space $Y$ as a finer relation, and as a consequence, this space $Y$ has a relation $R$ such that $Y/R \approx Y$. For example, if you collapse a longitude from the north pole to the south pole to a point you still have a sphere.
