The other day, a friend of mine told me that he found this exercise on web:
Prove that we can obtain a Torus from the quotient of a Sphere.
And I'm not sure this is possible. First of all, the sphere $S^{2}$ with a handle is the torus $\mathbb{T}$, so the quotient should be something like this. I have seen here that I can't find a continuous open surjection from $S^{2}$ to $\mathbb{T}$. Now my question is: Is it possible to find a closed continuous surjection? Because if not, this implies that we can't find any quotient homeomorphic to $\mathbb{T}$.
Note that continuous maps may be neither closed nor open.
It is possible to exhibit the torus as a quotient of a sphere. Here is an explicit map that does this:
Take the maps $$(x,y,z)\in S^2\mapsto \left(\cos 2\pi y,\sin 2\pi y\right)\in S^1$$
and
$$(x,y,z)\in S^2\mapsto \left(\cos 2\pi z,\sin 2\pi z\right)\in S^1$$
They are continuous, and so their product is a continuous map $S^2\to S^1\times S^1$. It is also clearly surjective.
Correspondingly, there is an equivalence relation $\sim $ on $S^2$ with $(x,y,z)\sim (x',y',z')$ iff $x\equiv x'\bmod 1$ and $y\equiv y'\bmod 1$, and $S^1\times S^1\cong S^2/\sim$.