First the question, and then my work, and finally, my explicit question. The exercise is from the book An Intoduction to Algebraic Topology by Andrew Wallace.
Chapter IV.3 Exercise 8. Let $f$ be a closed path on a sphere $S$, based on a point $x$. Prove that $f$ is homotopic with respect to the base-point $x$ to a product, $f_1f_2f_3\dots f_r$ where each $f_i$ is a homeomorphism of $I$ onto some arc of a great circle of $S$.
There is also a hint:
Hint: $S$ can be covered by two open sets $U$ and $V$ each homeomorphic to an open circular disc; use the continuity of $f$ to subdivide $I$ into intervals $I_1\text{, }I_2\text{, }\dots\text{, }I_r$ such that each $f(I_i)\subset U$ or $V$. Hence show $f$ homotopic to a product $g_1g_2\dots g_r$ each $g_i$ being an open path on $U$ or $V$. Apply Exercise 5 above to replace each $g$ by an arc of a great circle.
Exercise 5 says:
Chapter IV.3 Exercise 5. Let $E$ be a simply connected space and let $x$ and $y$ be two points of $E$. Prove that any two paths $f$ and $g$ in $E$ joining $x$ and $y$ are homotopic with respect to the fixed end-points $x$ and $y$.
The hint suggests that we proceed in the following way:
- Find $U$ and $V$ which cover $S$ and are homeomorphic to an open circular disk.
- Find a finite partition $I$ such that the image of each subinterval lies wholly within either $U\text{ or }V\text{.}$
- Show that $f$ is homotopic to a product of (open) paths on $U\text{ or }V\text{.}$
- Immediately conclude that each $g_i$ is homotopic with respect to its end-points to an arc of a great circle.
1. $U$ and $V$ can be found by taking overlapping interiors of spherical caps and using stereographic projection to demonstrate homeomorphism.
2. I am having difficulty with the "finite" part of this step. If $f$ is a space-filling loop, it can pass the frontiers of $U\text{ and }V$ infinitely many times, mandating that any such family of subintervals whose images lie wholly within $U\text{ or }V$ must be infinite.
3. I have not finished this step to my satisfaction, but my sketch of it follows. Let $I_1,I_2,\dots,I_r$ be the subintervals whose images under $f$ lie wholly in $U\text{ or }V$. Define these images as $f_{I_1},f_{I_2},\dots,f_{I_r}$ and note that these are restrictions of the function $f$ to each subinterval. Then, we can define functions $g_1,g_2,\dots,g_r$ which map the interval to each respective subinterval, and then follow the restriction defined on that subinterval to the sphere. Each restriction is continuous, so each $g_i$ is continuous, so each $g_i$ is a path joining the end-points of its associated subinterval lying wholly within either $U\text{ or }V\text{.}$
4. $U\text{ and }V$ are both homeomorphic to a disk, which is simply connected. Then, as each $g_i$ lies wholly within either, its image under composition of the homeomorphism is also a path. Then, that image is homotopic with respect to its endpoints to the straight line between the endpoints, which maps directly to an arc of a great circle.
Here are my questions.. 1. How do I go about showing step 2 from the hint as stated above? 2. Does the compactness of the closed loop or the sphere have anything to do with these homotopies? 3. Is there any other issues I'm missing from my understanding of this exercise?
Your "spherical caps" should be larger than a hemisphere. I tend to like the northern cap $U$ (everything north of and including the tropic of capricorn) and the southern cap $V$ (everything south of and including the tropic of cancer).
For 2: you've got a continuous map from a compact set (the interval $I$) into $S^2$. The preimages of $U$ and $V$ are open sets (probably not connected) that cover all of $I$. Hence the individual components of those sets, all taken together, form an open cover of $I$. Since $I$ is compact, finitely many of them also cover $I$.
Your argument about "crossing the border" is not quite right: to leave $U$, the curve must go below 23 degrees south; to leave $V$, it must go north of 23 degrees north. It can do this finitely many times while still crossing the boundary of $U$ (while not leaving $V$) or the boundary of $V$ (but not leaving $U$) infinitely many times. (For further details of this not-so-clear argument, see the comments.)
So to answer your question, "Yes, compactness matters." But it only matters for getting a finite collection of endpoints to work with. After that, it's just retraction onto the great arc between any successive pair of endpoints, followed by a "within the arc" homotopy to a direct path from one end to the other rather than a path that goes forward, then back, then forward some more, then stands still, etc.