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.

Solution 1:

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.

Solution 2:

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 $\mathbb{S}^2$.