Suppose we have a riemmanian manifold $(\mathcal{M}, g)$. Let $\mathcal{N}$ be properly embedded submanifold of $\mathcal{M}$ with induced metric $\bar{g}$ from $g$. ($\mathcal{N}$ is not necessarily totally geodesic manifold.)

Let $a, b \in \mathcal{N} \subset \mathcal{M}$ be two distinct points. Say $\gamma(t)$ is unique geodesic curve on $\mathcal{M}$ with metric $g$ and $\bar{\gamma}(t)$ is also unique geodesic curve on $\mathcal{N}$ with metric $\bar{g}$ with $\gamma(0) = \bar{\gamma}(0) = a$ and $\gamma(1) = \bar{\gamma}(1) = b$.

Question : If $c(t) \in \mathcal{N}$ is the closest point from $\gamma(t)\in\mathcal{M}$ to submanifold $\mathcal{N}$, then does $c(t)$ always belong to geodesic curve $\bar{\gamma}$?

Actually, this is just my guess, since it holds true for $\mathcal{M} = R^{3}$, $\mathcal{N} = S^{2}$. Is it generally true? or for what condition does it hold true? Curious whether there is some relevant theorem regarding this...

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    $\begingroup$ Of course, this is false if $\mathcal N$ is itself a non-geodesic curve. I'm thinking about a more substantive counterexample. $\endgroup$ – Ted Shifrin May 19 '17 at 22:40
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    $\begingroup$ @Ted: It's badly false: Let $a$ and $b$ be near-antipodes on a sphere, $\bar{C}$ the great circle arc from $a$ to $b$, $C = ab$ the Euclidean segment, and create an execrescence from the sphere that misses $\bar{C}$ (and so doesn't change the notion of geodesic) but "closely envelopes" $C$ (so the closest points of $N$ to points of $C$ are nowhere near $\bar{C}$). :) $\endgroup$ – Andrew D. Hwang May 19 '17 at 23:01

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