${\textbf{Thm: }}$ Suppose ${(M,g)}$ is a complete manifold

$\displaystyle \text{Sec}_M\leq 0$

Then ${\text{exp}_p:T_pM\rightarrow M}$ is a covering map. Moreover if ${M}$ is simply connected, then ${M}$ is diffeomorphic to ${\mathbb{R}^n}$.

Remark: ${p}$ is the north pole of ${\mathbb{S}^n}$, ${\text{exp}_p:\mathbb{R}^n\rightarrow \mathbb{S}^n}$ is the exponential map, then ${\text{exp}_p^*g}$ will not be a metric on ${\mathbb{R}^n}$ and ${\text{exp}_p^{-1}(-p)}$ are concentric circles. Proof: Since ${\text{Sec}_M\leq 0}$, we have no conjugate points of ${p}$, ${d\text{exp}_p}$ is nonsingular on ${T_pM}$. So ${\tilde{g}=\text{exp}_p^*g}$ will be a metric on ${M}$. In fact ${(T_pM,\tilde{g})}$ is a complete manifold. This is implied from Hopf-Rinow thm and the fact that lines through origin of ${T_pM}$ are geodesics under ${\tilde{g}}$ and they can be extended to infinity.

Since ${\text{exp}_p}$ is a local isometry between ${(T_pM,\tilde{g})}$ and ${(M,g)}$, then

$\displaystyle \{V_i|\text{exp}_pV_i=q\}$

is a discrete set. Choose ${0<\delta<\text{injrad}(q)}$. Denote ${\text{exp}_p=\phi}$

$\displaystyle \phi^{-1}(B_\delta(q))=\cup_iB_{\delta}(V_i)$

$\displaystyle \phi: B_{\delta}(V_i)\rightarrow B_\delta(q)\text{ is homeomorphism}$

$\displaystyle B_i\cap B_j=\emptyset,\quad i\neq j$

namely ${\phi}$ will be the covering map.

For any ${p'\in B_\delta(q)}$, there exists a geodesic ${\gamma}$ connects ${p'}$ and ${q}$. For any ${W_i\in \text{exp}^{-1}_pp'}$, there exists a local lifting of ${\gamma}$, say ${\tilde{\gamma}}$, such that ${\gamma=\phi\circ\tilde\gamma}$. Since ${\phi}$ is a local isometry, then it is length preserving. Since ${\tilde{\gamma}}$ ends up at some ${V_{i_0}}$, then ${W_i}$ must be in ${B_\delta(V_{i_0})}$.

Next ${\phi:B_{\delta}(V_i)\rightarrow B_\delta(q)}$ is injective. Otherwise suppose ${W_1, W_2\in B_{\delta}(V_i)}$ with ${\phi(W_1)=\phi(W_2)=p'}$. Suppose ${\tilde\gamma_1}$ and ${\tilde\gamma_2}$ are geodesics connecting ${V_i}$ and ${W_1}$ and ${W_2}$ respectively, then both ${\gamma_1=\phi\circ\tilde\gamma_1}$ and ${\gamma_2=\phi\circ\tilde\gamma_2}$ will connect ${q}$ and ${p'}$. ${\gamma_1}$ and ${\gamma_2}$ will be two different geodesics because ${\phi}$ is a local isometry. This can not happen because ${p'}$ is inside the injective domain.

${\phi:B_{\delta}(V_i)\rightarrow B_\delta(q)}$ is onto. Suppose ${\text{exp}_qtv}$ be the minimal geodesic from ${q}$ to ${p'}$. Then there exist ${w\in T_{V_i}T_pM}$ such that ${d\phi (w)=v}$, then ${\phi(\text{exp}_{V_i}tw)}$ will be a geodesic in ${M}$ with initial speed ${v}$. Therefore it must concide with ${\text{exp}_ptv}$. Since ${\phi}$ preserve the length, we have ${p'\in (B_\delta(V_i))}$. ${\phi}$ is onto.

Suppose ${w\in B_\delta(V_i)\cap B_\delta(V_j)}$, ${\tilde{\gamma}_i}$ and ${\tilde\gamma_j}$ are two minimal geodesics from ${w}$ to ${V_i}$ and ${V_j}$. Then ${\tilde\gamma_i}$ and ${\tilde\gamma_j}$ are different near ${w}$, then ${\gamma_i=\phi\circ\tilde{\gamma}_i}$ and ${\gamma_j=\phi\circ\tilde{\gamma}_j}$ will be two different geodesics connecting ${q}$ and ${\phi(w)}$. Contracdtion implies ${B_\delta(V_i)}$ and ${B_\delta(v_j)}$ are disjoint. $\Box$

Remark: If we change the restriction to ${\text{Ric}_M\leq 0}$, the theorem will not be true. Counter example comes from the Calabi-Yau manifold, which is compact, simply connected and has vanishing ricci curvature.