I am working on my own version of a proof of the Jordan Separation Theorem (just for fun - I know it's been proved countless times) and in the course of so doing I use the apparently fairly obvious result that $S^n$ with two distinct points removed is homotopically equivalent to $S^{n-1}$.

In trying to prove that result I find that I need to show that the maximum distance between a geodesic segment (great hypercircle) from P to x and a geodesic segment from P to Q (P and Q not anitpodal to one another) is a continuous function of x, at Q. That is, I want to prove that, $\forall P,Q\in S^n, P\notin \{Q,Q'\}, \lim_{x\to Q}\textrm{gdist}(x)=0$, where $$\textrm{gdist}(x)\equiv \sup\{\inf \{d(z,y)\vert z\in \bar{PQ}\}\vert y\in \bar{Px}\}$$

$Q'$ indicates the antipodal point to $Q$ and $\bar{AB}$, for non-antipodal points A and B, denotes the path of the shortest geodesic curve from A to B. Such a path will be unique, since A,B are not antipodes of one another.

I am having trouble proving this because it is a global property that I am trying to deduce from local phenomena. That is, I am trying to show that shifting the endpoint of a geodesic segment by a tiny bit can only produce a tiny separation of the perturbed segment from the original one, all the way along the segment.

I have a feeling that this result should be true generally for smooth Riemannian manifolds, provided there is a unique shortest geodesic segment from P to Q (generalising the concept of 'non-antipodal' points to non-spheres).

I also have a feeling that this is probably a standard theorem of differential geometry. Yet, reviewing the main theorems I have learned (I mostly use John Lee's 'Riemannian Manifolds: An Introduction to Curvature'), I can't see such a result, or one that easily leads to it.

Any suggestions would be greatly appreciated - whether just for the hyperspherical case or the general Riemannian one. Thank you.

[Edited 26 Mar]Here is some info on why I need this answer, in response to the comment of Jesus RS below: The reason I need the proof is in order to show that one can smoothly transform $S^n$ with any two distinct points $a_1,a_2$ removed into $S^n$ with two antipodal points $P_1,P_2$ (poles) removed, where $a_i$ is in the hemisphere of $P_i$ for $i\in\{1,2\}$. The homotopy I developed to do this is as follows:

Let $\zeta_i:I\to S^n$ be the constant-speed parametrisation of the shortest geodesic segment from $a_i$ to $P_i$.

For each $x$ in the hemisphere of $P_i$ let $\xi_x$ be the first point on $\mathscr{C}$, the equator dividing the two hemispheres, encountered by the geodesic that starts at $a_i$ and then passes through $x$ before reaching $\mathscr{C}$.

Then let $\gamma_i^{x,t}:I\to S^n$ be the constant-speed parametrisation of the shortest geodesic segment from $\zeta_i(t)$ to $\xi_x $, and set $\alpha_x\equiv (\gamma_i^{x,0})^{-1}(x)$.

Note that the paths Im$\,\zeta_i$ and Im$\,\gamma^{x,t}_i$ will each be less than half a great circle and, except for the final point of $\gamma^{x,t}_i$, will lie entirely within the hemisphere of $P_i$.

Then I believe the map $G:(S^n\smallsetminus \{a_1,a_2\})\times I\to S^n$, given by $G(a_i,t)=\zeta_i(t)$ and $G(x,t)=\gamma_i^{x,t}(\alpha_x)$ if $x\neq \alpha_i$, is a homotopy that transforms $S^n\smallsetminus \{a_1,a_2\}$ to $S^n\smallsetminus \{P_1,P_2\}$.

To prove that is correct, I need to prove that $G$ is continuous, which requires the result stated above, because I need to show that $\forall \epsilon>0\forall t\forall x\exists \delta>0$ such that $d(\gamma_i^{x,t}(\alpha_x),\gamma_i^{x,t+\delta}(\alpha_x))<\epsilon$.

As an aside, I have found a way around this problem by finding a homotopy that directly transforms $S^n\smallsetminus \{a_1,a_2\}$ to the equator $\mathscr{C}$, which is easily shown to be homeomorphic to $S^{n-1}$. That avoids the necessity of going via $S^n\smallsetminus \{P-1,P_2\}$. Nevertheless I would still like to prove that $G$ is a homotopy.

  • $\begingroup$ Is that something wrong with your definition of gdist? Now $y\in \overline{Py}$ and so gdist$(x)= 0$ for any $x$? $\endgroup$
    – user99914
    Feb 25 '15 at 1:46
  • $\begingroup$ Thanks John. Yes my $\bar{Py}$ should have been $\bar{PQ}$. I have corrected it above. $\endgroup$ Feb 25 '15 at 1:59
  • $\begingroup$ Your assertion is correct. The reason is that if that unique geodesic between $P$ and $Q$ is given by $\exp (tv)$, then for any $x$ close to $Q$, there is also a length minimizing geodesic joining $P$ to $x$, and is given by $\exp (tv_x)$ for some $v_x$ close to $v$. $\endgroup$
    – user99914
    Feb 25 '15 at 7:05
  • $\begingroup$ OK, that breaks the problem down into two lemmas (1) $\forall P,Q \forall \epsilon>0\exists\delta>0$ such that $x\in B_\delta(Q)\Rightarrow $ still editing this, please don't read yet $\endgroup$ Feb 25 '15 at 10:14
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    $\begingroup$ You should be able to deduce this using the fact that the exponential map is a diffeomorphism when restricted to the interior of the segment domain. Once you have fixed $Q$ you can restrict to a precompact neighbourhood of $\overline{PQ}$, on which you will have a uniform bound on $D \exp$. $\endgroup$ Feb 27 '15 at 1:36

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