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I want to prove that the sphere $S^2$ is simply connected. The following argument appeared at Munkre's book on topology and it is "flawed", since there are surjective loops in the sphere $S^2$. But I heard that it is possible to show that such a surjective loop is homotopic to a non-surjective loop and hence the argument can be used again to finally prove that $S^2$ is simply connected.

The argument is the following:

Let $f:I = [0,1] \rightarrow S^2$ be a loop in $S^2$. Assuming that $f$ is not surjective, let $p\not\in f(I)$. Now considering $\varphi: S^2 \setminus\{p\} \rightarrow \mathbb R^2$, the stereographic projection at $p$, we obtain a loop in $\mathbb R^2: \varphi \circ f.$ Since $\mathbb R^2$ is simply connected, we obtain that $\varphi\circ f$ is path homotopic to a constant path given by the path homotopy $H$. Hence, $\varphi^{-1}\circ H$ is a path homotopy in $S^2$ between $f$ and a constant path.

Now the question is: is it true that a surjective loop in $S^2$ is path homotopic to a non-surjective loop in $S^2$? How to prove that? Thank you!

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    $\begingroup$ Show it's homotopic to a loop made up of great circle arcs. $\endgroup$ Commented Jun 27, 2019 at 17:55
  • $\begingroup$ It is of course true that such a homotopy exists, but a much stronger result is usually proven which says that any map between CW complexes is homotopic to a cellular map. $\endgroup$ Commented Jun 27, 2019 at 17:56
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    $\begingroup$ You can use PL-approximation such as this lemma 4.10 in Hatcher: pi.math.cornell.edu/~hatcher/AT/AT4.10rev.pdf (There is also smooth approximation.) Both ideas lead to a non-surjective map in the same homotopy class. $\endgroup$
    – user17892
    Commented Jun 27, 2019 at 18:41
  • $\begingroup$ I'm curious, didn't the textbook go on to discuss the case that $f$ is surjective? $\endgroup$
    – Lee Mosher
    Commented Jun 27, 2019 at 19:23
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    $\begingroup$ Let $p\in S^2$, and let $U=S^2-p$ and $V$ be a small open disk about $p$, so $U$ and $V$ together form an open cover. Let $x\in U\cap V$ be the basepoint. If you follow the proof of the van Kampen theorem carefully, you can see how to break $f$ up using compactness, homotope it to a composition of loops at $x$, then contract all the loops that are in $V$, resulting in a loop that avoids $p$. (This works for any manifold, not just $S^2$.) $\endgroup$ Commented Jun 27, 2019 at 20:57

1 Answer 1

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It's just a Lebesgue number argument. I thought this argument would be in Munkres "Topology", he is usually good on Lebesgue number arguments, but I don't have my copy to check.

Anyway, here is a version of the argument.

Let $\{U_i\}$ be the open cover of $S^2$ by open hemispheres. Using the given loop $f : [0,1] \to S^2$ we obtain an open cover $\{f^{-1}(U_i)\}$ of $[0,1]$. Let $\lambda>0$ be a Lebesgue number for the latter open cover. Choose an integer $n \ge 1$ such that $\frac{1}{n} < \lambda$. Subdivide the interval into subintervals of length $\frac{1}{n}$, $$[0,1] = [x_0,x_1] \cup [x_1,x_2] \cup \ldots \cup [x_{n-1},x_n] $$ It follows that for each $i=1,...,n$ the path $f| [x_{i-1},x_i]$ is contained in an open hemisphere of $S^2$. Let $\delta_i$ be the unique great circle path in that open hemisphere having the same endpoints as $f | [x_{i-1},x_i]$. Since open hemispheres are homomorphic to $\mathbb R^2$, it follows that $f | [x_{i-1},x_i]$ is path homotopic to $\delta_i$. Thus $f$ is path homotopic to the concatenation $\delta_1 * \ldots * \delta_n$.

Now convince yourself that a union of finitely many great circle subpaths of $S^2$ is not surjective in $S^2$.

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    $\begingroup$ Five months later, I've looked up Munkres "Topology", 2nd edition. He proves simple connectivity of $S^n$ ($n \ge 2$) in Corollary 59.2. His proof is indeed a very similar Lebesgue number argument, although for pedagogical purposes he puts Lebesgue number portion of the argument into a more general result (Theorem 59.1) from which Corollary 59.2 is then derived. $\endgroup$
    – Lee Mosher
    Commented Nov 13, 2019 at 19:21

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