Define an equivalence relation on $S^1\times[0,1]$ by: $(x,t)$~$(y,s) \iff xt=st$. Show that $(S^1\times [0,1])/$~ is homeomorphic to the unit disc $D^2$.
My attempt: Let $g: S^1\times[0,1]\to D^2$ by $g(x,t)=xt$. Now $g(x,t)=g(y,s)\iff xt=ys\iff (x,t)$~$(y,s)$. This shows two things: that $g$ is injective (forward direction) and $g$ is constant on each equivalence class of $S^1\times [0,1]$ (reverse direction). The latter means that $g$ induces a map $f:(S^1\times [0,1])/$~$\to D^2$. Now $S^1$ and $[0,1]$ are compact, so $S^1\times [0,1]$ is compact, and so $(S^1\times[0,1])$/~ is compact by the projection map $\pi$. We know $g$ is a surjection as well because for $y\in D^2$, we have $g(y/\|y\|,y)=y$. Since $g=f\circ\pi$, we have that $f$ is bijective, and we have that $f$ is continuous by the universal mapping property of quotients. Hence, $f$ is a continuous bijection from a compact space to a Hausdorff space, so it is also a homeomorphism.
Does anyone see any problems with my argument?
