Proposition: If a Riemannian manifold $(M,g)$ has non-positive sectional curvature, then $\forall p\in M,\mathrm{conj}(p)=\emptyset$.

Proof: Suppose $q\in\mathrm{conj}(p)$ is conjugate to $p$ along geodesic $\gamma:[0,1]\to M$ by Jacobi field $J$ such that $J(0)=J(1)=0$.

Consider $\varphi(t)=||J(t)||^2$.

$\begin{aligned}\text{Then }\varphi^{\prime\prime}(t)=&\left\langle\hat{\nabla}_{\frac{\mathrm{d}}{\mathrm{d}t}}\hat{\nabla}_{\frac{\mathrm{d}}{\mathrm{d}t}}J,J\right\rangle+\left\langle\hat{\nabla}_{\frac{\mathrm{d}}{\mathrm{d}t}} J,\hat{\nabla}_{\frac{\mathrm{d}}{\mathrm{d}t}} J\right\rangle\\=&\langle-R(J,\gamma^\prime)\gamma^\prime,J\rangle+\left|\left|\hat{\nabla}_{\frac{\mathrm{d}}{\mathrm{d}t}}J\right|\right|^2\\=&-R(J,\gamma^\prime,\gamma^\prime,J)+\left|\left|\hat{\nabla}_{\frac{\mathrm{d}}{\mathrm{d}t}}J\right|\right|^2\\\geqslant&0\end{aligned}$

And since $\varphi(t)\geqslant 0,\varphi(0)=\varphi(1)=0$.

So $\varphi\equiv 0$, i.e. $J$ is constant, contradiction!

Corollary: If a complete Riemannian manifold $(M,g)$ has non-positive sectional curvature, then exponential map $\mathrm{exp}_p:T_pM\to M$ is local diffeomorphic for any $p\in M$.

Remark: the exponential map $\mathrm{exp}_p$ is actually a covering map for any $p\in M$.