Le $(M,g)$ be Riemannian manifold. Fix point $p\in M$. We can define the map

$$\exp: U \subset T_p M \rightarrow M$$

$$\exp(X) = \gamma_{p,X}(1)$$

where $t\mapsto \gamma_{p,X}(t)$ is geodesic such that $\gamma_{p,X}(0)=p$, $\frac{d}{dt}\gamma_{p,X}(t)=X$. The set $U \subset T_p M$ is open neighbourhood of $0$ and is chosen such that the map $\exp$ is injective. One can show that for every metric $g$, there is some open $U$. Thus the map $\exp$ parametrizes some neighbourhood of point $p\in M$.

Fix $M=\mathbb{R}^3$ and define differentiable one parameter family of smooth metric tensors

$$[0,1]\ni s\mapsto g(s)=g_0 + s \delta g +O(s^2)$$

Obviously U and $\exp$ depend on $g$, so we might call them $U_g$, $\exp_g$.

The question: Is there open set $O\subset U_{g_0}\subset T_p M$ such that for every differentiable one parameter family of metric tensors and all $s \leq s_{\textrm{max}}$ ($s_{\textrm{max}}$ might depend on the choice of family) $O\subset U_{g(s)}$.

Intuitively it means that if $X \in O$ we can change infinitesimally $g$ in all possible directions $\delta g$ such that $\exp_g(X)$ makes sense. In particular it should be possible to determine the derivative of $\exp_g(X)$ with respect to $g$, i.e. the map $\delta g \mapsto \frac{d}{ds} \exp_{g(s)}(X)$, where as before $g(s)=g_0 + s \delta g +O(s^2)$. If the answer to the question is negative then the derivative might be defined only in some directions $\delta g$.

willdepend on the family. This is true by the usual compactness arguments (something like continuity of solutions to differential equations w.r.t. parameters). $\endgroup$ – Alex Degtyarev Mar 15 '14 at 10:44