# Proving directly that $S^2$ is simply connected: is a surjective loop homotopic to a non-surjective one?

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!

• Show it's homotopic to a loop made up of great circle arcs. Commented Jun 27, 2019 at 17:55
• 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. Commented Jun 27, 2019 at 17:56
• 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.
– user17892
Commented Jun 27, 2019 at 18:41
• I'm curious, didn't the textbook go on to discuss the case that $f$ is surjective? Commented Jun 27, 2019 at 19:23
• 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 $S^2$.) Commented Jun 27, 2019 at 20:57

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 $$S^2$$.
• 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. Commented Nov 13, 2019 at 19:21