Bounding the injectivity radius from below.

Question

Let $(M, g)$ be a finite-dimensional Riemannian manifold, and let $S \subseteq M$ be a compact set.

I want to prove the following fact:

There exists a number $\epsilon > 0$ such that the exponential map $\exp_p : B_{\epsilon}(0) \rightarrow M$ at $p$ is a diffeomorphism onto an open set, for all $p \in S$.

Now, I actually have a proof that involves considering the map $E : TM \rightarrow M \times M$ given by $E(v) = (\pi(v), \exp_{\pi(v)}(v))$ (when $M$ is complete; otherwise $E$ is defined on an open subset of $TM$ containing the zero section), showing that it is a local diffeomorphism near the zero section and arguing by contradiction.

However, I don't like this proof very much, and I have two related questions:

Question 1: Is there an easier way to establish this fact?

Question 2: What happens if $M$ is infinite-dimensional?

Thanks.

Answer 1

Let me adress Question 1 (if $M$ has infinite-dimension many things may happen, and you should be more specific about what you consider).

1. By the inverse function theorem, for ever $p \in S$ there exists some $\epsilon > 0$ such that $\exp_p : B_{\epsilon} (0) \rightarrow M$ is a diffeomorphism. Denote the image of this map $U_{\epsilon}(p)$.
2. Since $S$ is compact, there are finitely many points $p_i \in S$ with corresponding $\epsilon_i > 0$ and $S \subset \bigcup U_{\epsilon_i}(p_i)$. Moreover, there exists a $\delta > 0$ such that for every $q \in S$ the neighbourhood $U_{\delta}(q)$ is contained in one of the $U_i$ ($\delta$ is just the Lebesgue-number of the covering)
3. Convince yourself, that $exp_q : B_{\delta}(q) \rightarrow U_{\delta}(q)$ is a diffeomorphism for every $q \in S$ using the previous step.

This prove should be more intuitive, but to finish the last step in full detail requires some additional work.