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