Does there exists a continuous closed surjection from $S^{2}$ to $S^{1}\times S^{1}$

Question

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 $\mathbb{T}$.

Joshua Tilley · Accepted Answer · 2023-12-12T01:11:42.050

3

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 $S^1\times S^1\cong S^2/\sim$.

edited Dec 12 '23 at 01:11

answered Dec 11 '23 at 11:58

Joshua Tilley

7,070

2

Note that all maps between compact Hausdorff spaces are closed. – Tyrone Dec 11 '23 at 14:12
I didn't know that! Thanks for the correction. – Joshua Tilley Dec 11 '23 at 14:16
@JoshuaTilley I know that a continuous map may be neither closed or open. However, if it is neither closed or open, we can't prove that the quotient is homeomorphic. Anyway, as Tyrone said, in this case the map is closed. – Dungessio Dec 11 '23 at 20:22
@JoshuaTilley However, thanks! – Dungessio Dec 11 '23 at 20:23
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Related: Any surjective continuous function from a compact space to a Hausdorff space is a quotient map. – Lee Mosher Dec 12 '23 at 02:45
@LeeMosher sure. I realized later that I had only to find a continuous surjective map since by the properties of the spaces, it shall be autmatically closed. – Dungessio Dec 12 '23 at 08:29

Does there exists a continuous closed surjection from $S^{2}$ to $S^{1}\times S^{1}$

1 Answers1