# I cannot understand an explanation why 2-sphere is simply connected.

I am studying Elementary Differential Geometry written by Barrett O'Neill.

In page 188, Chapter 4.7, there is an explanation why 2-sphere is simply connected.

The following is from the text :

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The 2-sphere $\Sigma$ is simply connected. Consider the following scheme of proof. Let $\alpha$ be a loop in $\Sigma$ at, say, the north pole of $\Sigma$. Pick a point q not on $\alpha$. For simplicity, suppose q is the south pole. Now let x be the homotopy under which each point of $\alpha$ moves due north along a great circle, reaching p in unit time. This x is a homotopy of $\alpha$ to a constant, as required.

But there is a difficulty here: finding the point q. In our usual case, where $\alpha$ is differentiable, techniques from advanced calculus will show that there is always a point q not on $\alpha$. However, if $\alpha$ is merely continuous, it may actually fill the entire sphere. In this case, topological methods can be used to deform $\alpha$ slightly, making it no longer space-filling: then the scheme above is valid.

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My questions are:

1. Why why why why does he pick q ??? for what ???

2. I cannot even grasp What he wants to say in the 2nd paragraph. Can you explain why the 2nd paragraph is needed?

Thank you.

• $\bf p$ should be $\bf q$. The second paragraph addresses the issue of whether such a $\bf q$ (a point not on $\alpha$) actually exists. – Anthony Carapetis Apr 9 '16 at 4:28
• Isn't a loop a 1-to-1 continuous image of $S^1$? – DanielWainfleet Apr 9 '16 at 4:30
• @user254665: No. Most importantly, a loop need not be 1-to-1. But also, the loop is the function itself, not the image. – Eric Stucky Apr 9 '16 at 5:37
• Instead of "due north", read "away from q towards p". The second paragraph talks about the fact that q (a point not on $\alpha$) doesn't always exist. Look up space-filling curves. – Akiva Weinberger Apr 10 '16 at 3:46
• @AkivaWeinberger :: Thanks for the answer. I am totally new to diffgeo and topology. As I understood, the homotopy x can deform the loop and make it shrink into a point. But why do we care "away from q" which is not on the loop while the deforming one is the loop? Do you mean if a loop is space-filling, for example, then there is not enough "extra rooms" to deform the loop? – DOTDO Apr 10 '16 at 14:49

Well, how would you extend the given homotopy across $q$? The set $S^2\setminus q$ is just $\mathbb{R}^2$ (consider, for example, the Riemann sphere); with that identification, the given homotopy is just the straight-line contraction $h_t(x) = tx$ of $\mathbb{R}^2$ to the origin. The space $S^2$ itself is certainly not contractible.
In brief, the argument above is that any loop $\alpha:S^1 \to S^2$ has to factor through the contractible space $\mathbb{R}^2$ and thus be null-homotopic. It only works if $\alpha$ is not surjective. As the text indicates, it can be eliminated even in the continuous category by, for example, taking a small contractible neighborhood $U$ around a point in $S^2$ and taking a homotopy of $\alpha\cap U$ to something reasonable.
• In a sense, but my point is that any loop $S^1 \to S^2$ can be written as a map $S^1 \to \mathbb{R}^2 \to S^2$ and thus is homotopy equivalent to a map $S^1 \to * \to S^2$. – anomaly Apr 10 '16 at 15:40