## A simple result on homotopy in Riemannian manifolds

This is a simple result on the homotopy of two curves with the same endpoints in a Riemannian manifold ${(M, g)}$, which is raised in today’s Riemannian geometry lesson. Professor sketched the proof of the result but there’s one point which seems unclear (at least to me), and I also have some doubt about the statement of the result. I am posting it here in the hope that some of you can actually help me improve the statement of the following result.

Theorem 1 Let ${\alpha: [0, 1]\rightarrow M}$ be a continuous curve, then there exists ${\epsilon>0}$ (depending on ${\alpha}$) such that if ${\beta:[0, 1]\rightarrow M}$ is another continuous curve with the same endpoints, satisfying

$\displaystyle \max_{t\in[0,1]}d(\alpha(t), \beta(t))<\epsilon. \ \ \ \ \ (1)$

Then ${\alpha}$ and ${\beta}$ are homotopic relative to their endpoints. i.e. there exists continuous ${\sigma=\sigma(s, t): [0, 1]\times [0,1]\rightarrow M}$ such that ${\sigma(0, t)=\alpha(t)}$, ${\sigma(1, t)=\beta(t)}$, ${\sigma(s, 0)=\alpha(0)=\beta(0)}$ for all ${s}$ and ${\sigma(s, 1)=\alpha(1)=\beta(1)}$ for all ${s}$.

Remark 1 In the lecture, it was said that ${\epsilon}$ can be chosen to be half of the injectivity radius of $M$ (if it is positive). I am not able to show this, and my ${\epsilon}$ cannot be explicitly written down (due to inverse function theorem argument, see below).

Proof: Let us define the map ${E: U\subset T(M)\rightarrow M\times M}$ by ${E(p, v)=(p, \exp_p(v))}$, where ${U}$ is an open set (which can be chosen to be maximal) containing the zero section ${0}$ of ${T(M)}$ such that the exponential map is well-defined. Then by inverse function theorem, ${E}$ is a diffeomorphism in an open set ${U'}$ of ${U}$, containing ${0}$. This is because ${E}$ is ${C^1}$ and the differential of ${E}$ at a “point” ${(p, 0)\in T(M)}$ is ${\begin{pmatrix} I &I\\ 0 &I \end{pmatrix}}$ which is invertible. (I’ve omitted some details. )

Let ${V=E(U')}$ which is an open set in ${M\times M}$, containing the diagonal ${\{(p, p)\in M\times M: p\in M\}}$. Let ${\epsilon}$ be so small that if ${\beta}$ satisfies (1) then ${(\alpha(t), \beta(t))\subset V}$ and by shrinking ${\epsilon}$ further, that ${\beta(t)}$ lies in the convex normal neighborhood of ${\alpha(t)}$ for all ${t}$ (i.e. every two points in this neighborhood can be joined by a unique minimal geodesic which is contained in this neighborhood). Then for all fixed ${t}$, there is a unique minimal geodesic ${\sigma_t=\sigma_t(s): [0, 1]\rightarrow M}$ such that ${\sigma_t(0)=\alpha(t), \sigma_t(1)=\beta(t)}$. Now we define ${\sigma(s, t)=\sigma_t(s)}$. Note that we then have ${\sigma(0, t)=\alpha(t)}$, ${\sigma(1, t)=\beta(t)}$, ${\sigma(s, 0)=\alpha(0)=\beta(0)}$ for all ${s}$ and ${\sigma(s, 1)=\alpha(1)=\beta(1)}$ for all ${s}$.

It remains to show that ${\sigma}$ is continuous (on the square ${[0,1]\times [0, 1]}$). By our construction, the inverse of ${E|_{U'}}$ (which for convenience still denote by ${E}$) exists on ${V}$. So let ${(\alpha(t), V(t))=E^{-1}(\alpha(t), \beta(t))}$, where ${V(t)\in T_{\alpha(t)}(M)}$, ${V(t)}$ will be continuous. It is then easy to see that

$\displaystyle \sigma(s, t)=\exp_{\alpha (t)}(s V(t))$

which is continuous. $\Box$