TITLE: Where is the exponential map a diffeomorphism?
QUESTION [14 upvotes]: Let $M$ be a closed compact Riemannian manifold. 
The exponential map $\mathrm{exp}:TM\to M\times M$ takes $(p,v)$ to $(p,\gamma_v(1))$, where $\gamma_v$ is the geodesic flow at $p$ in the direction of $v$. The exponential map $\mathrm{exp}_p:T_pM\to M$ is the projection to the second coordinate of the restriction of $\mathrm{exp}$.
By the inverse function theorem, $\mathrm{exp}$ is a diffeomorphism on a neighborhood of the zero section in $TM$. So in particular there is a value $\epsilon_{\mathrm{Diff}}$ which is the supremum of values so that $\mathrm{exp}$ restricted to disk bundles in $TM$ of radius less than $\epsilon_{\mathrm{Diff}}$ is a diffeomorphism.
On the other hand, the injectivity radius at $p$, $\epsilon_p$, is the supremum of $\epsilon$ such that $\mathrm{exp}_p$ is a diffeomorphism from the radius $\epsilon$ ball around $0$ in $T_pM$ onto its image. The injectivity radius of $M$ is $\epsilon_{\mathrm{inj}}=\min_p \epsilon_p$.
Clearly $\epsilon_{\mathrm{Diff}}\le \epsilon_{\mathrm{inj}}$. Is the converse true? Is that obvious? If not, what can go wrong? Is there a reference? I glanced at several books and couldn't find an explicit answer anywhere. In Berger's "A Panoramic View of Riemannian Geometry" there was something somewhat parenthetical along these lines but with no argument or reference.

REPLY [5 votes]: Yes.  It is obvious.  The cost is that any attempt to explain will probably take more words than necessary.  Here is one such attempt:
Suppose $\epsilon < \epsilon_{\mathrm{inj}}$, and consider the radius-$\epsilon$ disk bundle $\mathrm{T}^{<\epsilon}M \subseteq \mathrm T M$.  Then the restriction $\exp: \mathrm{T}^{<\epsilon}M \to M \times M$ is a map of bundles over $M$, where the latter is a bundle with projection being projection onto the first factor.  On each fiber it is the map $p$.  Consider local coordinates $(p,v)$, where $p$ is the coordinate on the base $M$ and $v$ restricts on each fiber to a coordinate on the fiber.  Your function is $(p,v) \mapsto (p, \exp_p(v))$.  The derivative of this function is of the form $\begin{pmatrix} 1 & * \\ 0 & \frac{\partial}{\partial v}\exp_p(v)\end{pmatrix}$.  You have assumed that $\frac{\partial}{\partial v}\exp_p(v)$ is invertible on $\mathrm{T}^{<\epsilon}M$, so your matrix is invertible, so your function is a local diffeomorphism.  But it is also an injection, since projection in the $p$ direction distinguishes the fibers, and $\epsilon$ is sufficiently small as to make it a fiberwise injection.
Thus $\epsilon_{\mathrm{inj}} \leq \epsilon_{\mathrm{diff}}$.

I can't resist mentioning one of my favorite facts from differential geometry.  Call a vector $v \in \mathrm T M$ focal if $\exp$ fails to be a local diffeomorphism at $v$.  The study of focal vectors is closely related to the study of the boundary value problem for geodesic flow.
As you point out, the focal vectors, if there are any, are not in a small neighborhood of the $0$ section $\mathrm T M$.  It turns out that if the metric is positive definite then along any ray in (a fiber of) $\mathrm T M$, the focal vectors are isolated.  But there are examples in Lorentzian signature where this fails.  In fact, my memory of the statement is the following.  Choose any closed subset $S \subseteq [0,1]$ that does not contain $0$.  Then there is a Lorentzian manifold $M$ and a ray $\{sv\}_{s\in \mathbb R_{\geq 0}} \subseteq \mathrm T M$ such that for $s\leq 1$, $sv$ is focal iff $s\in S$.  (This is about as strong as you could hope, since focality is clearly a closed condition.)  In particular, in Lorentzian signature, the set of focal vectors can have topology as bad as a Cantor set.
I thought I had references written down for these facts, but I can't seem to find them, sorry.  The fact about positive definite metrics is probably in Milnor's book, among other places.  The construction of Lorentzian metrics I found on arXiv by random googling, and I'm not at all confident that I'd be able to reproduce the experiment.

REPLY [2 votes]: It is a diffeomorphism on $\{X\in TM: \|X\|_g < \epsilon_{\pi_M(X)}\}$ which is open since $p\mapsto \epsilon_p$ is upper semicontinuous. This is a set which is an open ball in each fiber. In general, $\exp$ is a diffeomorphism on an even larger set (which need not be ball in each fiber). Obviously $\{X\in TM: \|X\|_g< \epsilon_{inj}\}$ is the largest "tube with constant radius" in that set.