A complete, simply connected Riemannian manifold has no conjugate points if and only if every geodesic is length-minimizing. I just realized that I don't know whether the same is true for a locally CAT(k) space (i.e. a geodesic space with curvature bounded above in the Alexandrov sense).


Thanks to Alexander and Bishop, there is a developed "geodesic analysis" in these spaces, including Jacobi fields and conjugate points. And there is a Cartan-Hadamard theorem for spaces without conjugate points: if it is simply connected, then every pair of points is connected by a unique geodesic.


In the Riemannian case, the converse statement follows from the basic fact that a geodesic beyond a conjugate point is no longer minimizing. This is proved by constructing a length-decreasing variation, or something similar, from a vanishing Jacobi field. Unfortunately, this argument uses a lower curvature bound. Well, not quite that, because it also works in Finsler geometry, but anyway it fails for CAT(k): on a bouquet of two spheres there are geodesics that remain minimizing beyond a conjugate point.


However this does not disprove the converse Cartan-Hadamard theorem. Hence the question:


Let $X$ be a space with curvature locally bounded above. Let's not talk about monsters: the space is complete, locally compact, all geodesics are extensible (otherwise one can play dirty tricks with a boundary). Suppose that every geodesic in $X$ is minimizing. Or even better: every pair of points is connected by a unique geodesic. Does this imply that the geodesics have no conjugate points?


**UPDATE**. Thanks to Henry Wilton, I've found that there is no standard definition of a conjugate point. In fact, some definitions are designed so as to imply the affirmative answer to my question immediately. When I asked the question, I meant the following (maybe not the best possible) definition.


Fix a point $p\in X$ and consider the space $X\_p$ of geodesic segments emanating from $p$. The segments are parametrized by $[0,1]$ proportionally to arc length. The space $X\_p$ is regarded with the $C^0$ metric. The exponential map $\exp\_p:X\_p\to X$ is defined by $\exp\_p(\gamma)=\gamma(1)$. A point $q=\gamma(1)$ is conjugate to $p$ along $\gamma$ iff $\exp\_p$ is not bi-Lipschitz near $\gamma$.



In a locally CAT(k) space, does uniqueness of geodesics imply the lack of conjugate points?
Consider a surface of revolution with an equator $\ell$ of lenght $2{\cdot}\pi$ such that its Gauss curvature $$K=1/\left(1+\sqrt[5]{\mathrm{dist}\_ \ell}\right).$$
Choose $z\in \ell$ and let $\Sigma=B\_{\pi/2}(z)$.
Clearly $\Sigma$ is a $\mathrm{CAT}(1)$-space it has just one pair of conjugate points (say $p$ and $q$ --- the ends of $\Sigma\cap\ell$) and it has unique geodesics between each pair.


It remains to make geodesics extensible. 
To do this, we take $\Lambda=(S^1\times [0,\infty), d)$ with flat metric and concave boundary $\partial \Lambda=\partial\Sigma$.
Then we glue $\Lambda$ and $\Sigma$ along the boundary.


The metric on $\Lambda$ is completely described by curvature $k(u)$ of its boundary [$u\in \partial \Lambda=\partial \Sigma$].
We only need to choose a function $k$ which is 


* on one had is large enough so that the glued surface still has unique geodesics between each pair (in particular $k(p)=k(q)=\infty$).
* on the other hand is $\int\_{S^1} k<\infty$, so that glued space is locally compact.


I believe it is possible...


