# Is it possible to obtain a sphere from a quotient of a torus?

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.

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

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).
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 $$\mathbb{S}^2$$.