Theorem: Let $(M^n,g)$ complete, simply connected Riemannian manifold with nonpositive sectional curvature. Then $M$ is diffeomorphic to $\mathbb{R}^n$.
Question: Can we change complete to "every two points can be joined by a minimizing geodesic"?
Theorem: Let $(M^n,g)$ complete, simply connected Riemannian manifold with nonpositive sectional curvature. Then $M$ is diffeomorphic to $\mathbb{R}^n$. Question: Can we change complete to "every two points can be joined by a minimizing geodesic"? 


Take a point $x \in M$ and consider the starshaped open set $U \subset T_xM$ where the exponential $$ exp_x : U \rightarrow M $$ is defined. $U$ is diffeomorphic to ${\mathbb R}^n$ and the map ${\rm exp_x}$ is (1) a local diffeomorphism (because of the absence of conjugate points = singularities of ${\rm exp}_x$); (2) surjective (because I can join any other point of $M$ to $x$ by a geodesic). I guess there is no problem and $M$ is diffeomorphic to $U$. Right?? Edit. As Claudio points out, the fact that ${\rm exp}_x$ is surjective and a local diffeomorphism does not imply that it is a covering map. Hence we cannot use the homotopy lifting property to conclude that it is injective and a diffeomorphism. 


Let $X$ be completion of $(M,g)$. Note that $X$ is simply connected. It follows since any loop in $X$ is a limit of a broken geodesic in $(M,g)$. Therefore $X$ is CAT(0), in particular any two points are joined by unique geodesic. Therefore $M$ is diffeomorphic to $\mathbb R^n$ as any starshaped domain (thanks to alvarezpaiva). 


You can't change it, but complete implies that there exists such a geodesic. http://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem but the conditions are NOT equivalent. Regard an open ball in the euclidean space. 

