10

I understand that a torus is obtained from a sphere by adding a handle. I'm working on a question which is asking if it is possible to obtain a sphere from a quotient of a torus? It seems like this should be possible by perhaps identifying the insides of the torus? But I'm not quite sure how to properly express this.

Help is very much appreciated.

Wooster
  • 3,775
  • 1
    Does quotient just mean surjective continuous map? Do you know the description of the torus as a square with opposite sides identified? If you identified those opposite sides to a point, I think you'll get a sphere. – CJD Jun 10 '14 at 13:17
  • That is what I was thinking, but do you actually get a sphere or just a point? – Wooster Jun 10 '14 at 13:18
  • See this question: http://math.stackexchange.com/q/809595/4583 – Ayman Hourieh Jun 10 '14 at 13:21
  • Wooster, you get a sphere, because the interior of the rectangle hasn't changed. This is topologically the same as taking a closed disk and identifying its boundary to a point, which likewise gives you a sphere. – Ted Shifrin Jun 10 '14 at 13:36
  • 4
    Every positive-dimensional compact manifold is homeomorphic to a quotient of any other compact positive-dimensional manifold. –  Aug 03 '15 at 18:39

2 Answers2

15

The answer is yes:

Consider the torus sitting in $\mathbb R^3$ like a donut on a table. Then you see that it is invarant by a rotation of $180$ degrees around an horizontal axis. The quotient by such involution is a sphere and the projection is wat is usually called a branched cover (with four branch points).

In general any orientd closed surface covers the sphere via a branched covering.

user126154
  • 7,551
11

Think of the torus $\mathbb{T}$ as the product of two circles: $\mathbb{T} = \mathbb{S}^1 \times \mathbb{S}^1$.

Define the figure-8 subset $E$ of $\mathbb{T}$ by $$E = \mathbb{S}^1 \times \{ p \} \cup \{ p \} \times \mathbb{S}^1,$$ where $p$ is any point of $\mathbb{S}^1$.

Then by identifying $E$ to a point, $\mathbb{T}$ becomes homeomorphic to $\mathbb{S}^2$.

It's easy to check that $\mathbb{T} - E$ is an open square $(0,1) \times (0,1)$. So the quotient space $\mathbb{T}/E$ is a closed square with its boundary identified to a point, hence homeomorphic to $\mathbb{S}^2$.

AmorFati
  • 1,888
Dan Asimov
  • 1,157