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.

  • 2
    $\begingroup$ Yes, that's exactly it. The northern and southern (closed) hemispheres of a sphere are topologically (closed) disks, intersecting on their boundaries. $\endgroup$ – Robert Israel Nov 18 '12 at 6:47
  • $\begingroup$ And you can visualize their common boundary as the equator of the sphere. $\endgroup$ – Brian M. Scott Nov 18 '12 at 7:28

Intuitively, you are gluing the two disks along their boundaries.

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, $$ \require{AMScd} \begin{CD} S^1 @>{\operatorname{inc}}>> D^2\\ @V{\operatorname{inc}}VV @VV{}V\\ D^2 @>>{}> S^2 \end{CD} $$


enter image description here

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.


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