# Every path in $S^n$ is path homotopic to a “polygonal” path

I recently have started learning some elementary homotopy theory, and I'm having some trouble proving this statement. Over here a "polygonal path" means a path consisting of a finite number of points $s_1 = 0 < s_2 < ...< s_n = 1$ and for our path $p$, for every $i$, $p(s_i)$ is joined by $p(s_{i+1})$ by some arc of a geodesic of $S^n$, i.e an arc of a great circle. How should I proceed?

-

Suppose $\gamma:[0,1]\to S^n$ is a continuous map. Let $\mathcal U$ be the open covering of $S^n$ by the eight open half hemispheres.

There is a positive integer $m$ such that for all $i\in\{0,\dots,m-1\}$ the image of the interval $[\tfrac{i}{m},\tfrac{i+1}m]$ under $\gamma$ is completely contained in one of the open sets in $\mathcal U$, which I will call $U_i$. This is a consequence of the fact that $[0,1]$ is a compact metric space, that $\{\gamma^{-1}(U):U\in\mathcal U\}$ is an open covering of $[0,1]$, and of the Lebesgue's number lemma.

Let now $\sigma:[0,1]\to S^n$ be the polygonal map such that

• $\sigma(\tfrac{i}{m})=\gamma(\tfrac{i}{m})$ for all $i\in\{0,\dots,m\}$, and

• for all $i\in\{0,\dots,m-1\}$, the restriction $\sigma:[\tfrac{i}{m},\tfrac{i+1}{m}]\to S^n$ is a geodesic, with image completely contained in $U_i$.

I leave it as an exercise for the OP to prove that $\sigma$ and $\gamma$ are homotopic paths (relative to their endpoints, even)

-

If $n=1$, then it would appear that any path must consist of a sequence of arcs. However, you could have a path that "turns around" infinitely many times, and thus not be polygonal. To deal with this, starting with a path $P$, form the path $Q$ that consists of $P$ followed by a single arc back to the starting point of $P$. Now we apply the result that the fundamental group of $S^1$ is $\mathbb{Z}$ to show that $Q$ is homotopic to a polygonal path. Then $P$ itself must be homotopic to a polygonal path.

For $n>1$, things are simpler. Here the fundamental group of $S^n$ is trivial (that is, $S^n$ is simply connected), and any path is homotopic to a single arc.

Edit: Consider how we might prove that any path in $S^n$, $n>1$, is homotopic to a single arc, without just appealing to a result about fundamental groups. We might as well just prove that any two paths with the same endpoint are homotopic. To start with, instead of spheres, look at $\mathbb{R}^n$. If we have two paths $P$ and $Q$ with $P(0) = Q(0)$, $P(1) = Q(1)$, we construct a homotopy as follows. For each $x \in [0,1]$, let $L_x : [0,1] \rightarrow \mathbb{R}^n$ be the line segment starting at $P(x)$ and ending at $Q(x)$. Then, the homotopy, $f : [0,1] \times [0,1] \rightarrow \mathbb{R}^n$, is given by $f(x, t) = L_x(t)$. In order to make this a complete proof, we need to show that $f$ is continuous. This is done by observing that $L_x(t) = (1-t) P(x) + t Q(x)$. Since $P$ and $Q$ are continuous, and products and sums of continuous functions are continuous, $f$ is continuous.

Now, let's turn back to paths $P,Q$ in $S^n$. Suppose we have a point $x \in S^n$ that does not lie on either path. Then, we are essentially done since $S^n$ with a point removed is homeomorphic to $\mathbb{R}^n$, and we took care of the latter space. It is, however, possible that no such $x$ exists. To fix this, pick $x$ so that it is not an endpoint of either path. Now we have to tweak our paths so that they do not pass through $x$. The argument showing that this can be done is contained in Proposition 1.14 of Hatcher's Algebraic Topology. We apply the argument to both $P$ and $Q$ to get paths that do not pass through $x$ but that are homotopic to $P$ and $Q$ respectively. (The path $f$ in the proposition is given to be a loop, but that does not affect this part of the argument.)

(Note that these arguments are, more or less, the arguments used to prove that the fundamental group is trivial in the first place.)

-
Since I'm just starting, I don't think I can use concepts such as the fundamental group but in fact I would like to do it using only the basic facts about path homotopies and the operations that are defined on them. –  Gromali Nov 7 '12 at 8:22
I think your claim is at least as "hard" as the results about the fundamental groups. Certainly, the assertion that any path is homotopic to a single arc is essentially equivalent to saying that the fundamental group is trivial. –  Eric M. Schmidt Nov 7 '12 at 8:42
No it is a rather simpler statement! –  Mariano Suárez-Alvarez Nov 7 '12 at 16:43
How so? If we know that a loop is homotopic to an arc, then the "arc" is a constant map, and we are done. –  Eric M. Schmidt Nov 7 '12 at 16:54
(I am refering to the claim by the OP that any path is homotopic to a polygonal, not yours that anypath is homotopic to a signle arc!) Notice that it is true that any curve in any Riemannian manifold, simply connected or not, is homotopic to a polygonal path in the sense of the OP —the argument is exactly the same as the one I gave in my answer, modulo that one needs to know there exist geodesically convex open sets. –  Mariano Suárez-Alvarez Nov 7 '12 at 16:58