# Why surface of sphere does not have global coordinate system?

I came across this statement in one of the introductions to differential geometry that some of the manifolds cannot be expressed with a single global coordinate system and one of the examples is the surface of a sphere.

A global coordinate system is one where we have one-to-one mapping from all points on manifold (S) to $$\mathbb R^n$$. I can map each point on a sphere to $$\mathbb R^3$$, hence I should have a global coordinate system?

Am I missing something very obvious?

• The dimension of the manifold must be the same as the dimension $n$ of $\Bbb R^n$ containing the image of the coordinate chart(s). Dec 16, 2019 at 4:32
• But then I can just say the dimension of surface of the sphere is 3? Dec 16, 2019 at 4:39
• No, the chart must be a local homeomorphism, which is why the above condition holds. Dec 16, 2019 at 4:44

A chart for a manifold $$M$$ ($$\dim M = n$$) is a pair $$(U,\varphi)$$ where $$U\subseteq M$$ is open and $$\varphi\colon U \to \varphi[U]\subseteq \Bbb R^n$$ is a homeomorphism onto the open set $$\varphi[U] \subseteq \Bbb R^n$$. If $$M$$ is compact, the existence of a global chart $$(M,\varphi)$$ means that the image $$\varphi[M] \subseteq \Bbb R^n$$ is open, non-empty, and compact (by continuity of $$\varphi$$). This is impossible.

• Thanks! I think I understand your argument and on reading further I figured it out too but it makes me wonder if there is a particular intuition or reason why we define the range space to be open set and not closed? Dec 16, 2019 at 4:42
• The same reason why when dealing with Calculus in general, one considers open sets: when computing the limit of a function as $x$ tends to some point $p$ in the open set, one can approach $p$ from all possible directions (as the open set contains some ball centered at $p$). And derivatives are, in the end of the day, limits. Our goal is to do Calculus on the manifold, by transferring everything from $\Bbb R^n$ via charts. Dec 16, 2019 at 4:44