# Surface of a 2-sphere expressed as union of two closed disks

I'm reading a First Course in Differential Geometry by Chuan-Chih Hsiung and on page 8 he says "A closed disk that is homeomorphic to $I^2$ [i.e. $I\times I$, where $I = [a, b]$] is connected. The surface $S^2$ of a 2-sphere can be expressed as the union of two closed disks with nonempty intersection."

I'm not sure what he means by the second sentence. Am I supposed to imagine two disks being deformed into the two halves of the sphere (so the disks touch each other at their circumferences)? I don't understand what it means to express the spherical surface as a union of two disks.

• Yes, that's exactly it. The northern and southern (closed) hemispheres of a sphere are topologically (closed) disks, intersecting on their boundaries. – Robert Israel Nov 18 '12 at 6:47
• And you can visualize their common boundary as the equator of the sphere. – Brian M. Scott Nov 18 '12 at 7:28

This can be made precise with the language of pushouts in $$\textbf{Top}$$ : Given $$f : X \to Y$$, $$G : X \to Z$$, the pushout of $$f$$ and $$g$$, denoted by $$Y \times_X Z$$ is defined to be $$Y \sqcup Z / \sim$$ where $$y \sim z$$ provided there exists $$x \in X$$ such that $$f(x) = y$$ and $$g(x) = z$$. Intuitively, you are gluing along the information given by $$f$$ and $$g$$.
In this particular situation, we have the following, $$\require{AMScd} \begin{CD} S^1 @>{\operatorname{inc}}>> D^2\\ @V{\operatorname{inc}}VV @VV{}V\\ D^2 @>>{}> S^2 \end{CD}$$
You can easily see that once you have the upper hemisphere (or lower) you can project onto the $$xy$$-plane by forgetting the $$z$$-coordinate. This is a homeomorphism, so taking the inverse allows you to replace $$D^2$$ with this "curved" disk, coinciding with the upper (or lower) hemisphere.