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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.

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    $\begingroup$ 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. $\endgroup$
    – CJD
    Jun 10, 2014 at 13:17
  • $\begingroup$ That is what I was thinking, but do you actually get a sphere or just a point? $\endgroup$
    – Wooster
    Jun 10, 2014 at 13:18
  • $\begingroup$ See this question: math.stackexchange.com/q/809595/4583 $\endgroup$ Jun 10, 2014 at 13:21
  • $\begingroup$ 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. $\endgroup$ Jun 10, 2014 at 13:36
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    $\begingroup$ Every positive-dimensional compact manifold is homeomorphic to a quotient of any other compact positive-dimensional manifold. $\endgroup$
    – user98602
    Aug 3, 2015 at 18:39

2 Answers 2

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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 what is usually called a branched cover (with four branch points).

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

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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 $\mathbb{S}^2$.

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