**Question:** Is every open convex subset $C$ of a Riemannian manifold $M$, necessarily contractible?


Here by a "convex subset" I mean a set $C$ having the property that between each pair of points in $C$ there is a unique geodesic contained in $C$. (For example an open hemisphere is convex inside the sphere)


I have tried to construct a deformation retract along the unique geodesic connecting every point in $C$ to a fixed point in $C$. But is this map always continuous? If the answer is not positive in general, I'm also interested in the special case of complete manifolds with non-positive curvature.


**Added.** For the case of complete manifolds with non-positive curvature one can argue as follows: Let's $q\in C$ be the fixed point and $p\in C$. Now if $\gamma(t) = \mathrm{exp}(tv) (t\in[0,1])$ is the unique geodesic in $C$ connecting $q$ to $p$, there is a neighborhood $U$ of $v$ in $T\_q M$ s.t. $\exp([0,1]\*U)$ is contained in $C$. Now since $\exp$ has no critical points in non-positive curvature, $\exp(U)$ contains an open neighborhood of $q$ and we can use inverse mapping theorem.


Thanks!



Is every open convex subset of a Riemannian manifold necessarily contractible?
Fix $q \in C$. The set $D\subseteq T\_q M$ of all tangent vectors $v$ such that $[0,1]\ni t \mapsto \exp\_q(tv)$ is the unique geodesic in $C$ connecting $q$ with $p$ is open and star shaped, hence contractible.


Then $\exp\_q: D \to C$ is a continuous map between manifods with the same dimension. It is also one-to-one, by your convexity hypothesis. By [Browuer invariance of domain](https://en.wikipedia.org/wiki/Invariance_of_domain) it is necessarily an homeomorphism.


