On a geodesic mapping of a square Let $X$ be a proper geodesic space which is uniquely geodesic. Let $\phi:[0,1]\times[0,1] \to X$ be a function which satisfies the following:
The maps $\phi(0,\cdot)$, $\phi(\cdot,0)$, $\phi(1,\cdot)$, and $\phi(\cdot,1)$ are all (linearly parametrized) geodesics. Furthermore, for each fixed $s$, the map $\phi(s,\cdot)$ is a (linearly parametrized) geodesic connecting $\phi(s,0)$ to $\phi(s,1)$.
Given the above conditions, is it true that that for any fixed $t$, the map $\phi(\cdot,t)$ is a geodesic connecting $\phi(0,t)$ to $\phi(1,t)$? If not, is there a condition we can apply for which this is true (e.g. the space must be Hadamard)?
 A: This is not true. Let $X$ be the unit sphere, or some hemisphere thereof, which we describe first in spherical coordinates.
Let $f(s,0)$ go east along the equator, $(\theta,\phi)=(2s\pi/3,\pi/2)$.
Let $f(s,1)$ go south from the North Pole, $(\theta,\phi)=(\pi,s\pi/3)$
Let $f(s,t)$ be $t$ of the way from $f(s,0)$ to $f(s,1)$.
Then $f(s,1/2)$ is not a geodesic.
Each $f(s,1/2)$ is the midpoint of $f(s,0)$ and $f(s,1)$, so it is proportional to  $f(s,0)+f(s,1)$ in $\mathbb{R}^3$. Thus in Cartesian coordinates:
\begin{align}
f\left(0,\frac12\right) \propto\, &
\big(\phantom{-\sqrt{3}}\,1\phantom{\sqrt{3}}, \ \ \ \ 0 \ \ ,\ \ 1\ \ \ \big) \\
f\left(\frac12,\frac12\right) \propto\, & \left(\phantom{-\sqrt{3}}0\phantom{\sqrt{3}},\ \frac{\sqrt{3}}2, \frac{\sqrt{3}}2\right)\\
f\left(1,\frac12\right) \propto\, &
\left(\frac{-1-\sqrt{3}}2, \frac{\sqrt{3}}2,\ \frac12\ \ \right)
\end{align}
These three vectors have non-zero determinant, so they are not in the same plane through the origin, and $f(1/2,1/2)$ is not on the geodesic between the other two.
