For sphere $S^2$, the conjugate locus of each any point in $S^2$ reduces to a single point. The problem is:

For sphere $S^2\subset\Bbb{R}^3$, the conjugate locus of each point in $S^2\subset\Bbb{R}^3$ reduces to a single point.

I can prove that any point is conjugate to it's antipodal point. But I can't prove that other points aren't conjugate with with each other.
Any reference or sketch of proof will be interesting.
Thanks in advance.

My tries: There's a proposition in my book (Differential Geometry of Curves and Surfaces Manfredo P. do carmo) which says:

PROPOSITION 5. Let $p, q \in S$ be two points of $S$ and let $\gamma ':[0,1] \to S$ be a geodesic joining $p = \gamma '(0)$ to $q = \exp_p(l\gamma '(0))$. Then $q$ is conjugate to $p$ relative to $\gamma$ if and only if $v = l\gamma '(0)$ is a critical point of $\exp_p:T_p(S) \to S$.

In this way I need calculate $(d \exp_p)_v(w)$ to observe for which $v,w$ we have $(d \exp_p)_v(w)=0$.
 A: Use the usual spherical coordinates $(\phi,\theta)$ such that $\theta \in S^1$ and $\phi \in [0,\pi]$ with $p=(0,0)$. We can turn these into a legitimate chart on the sphere minus a point using polar coordinates: if $(r,\vartheta)$ are the polar coordinates on $B_\pi$ (the open disc of radius $\pi$) then the chart is simply the "identity" $(\phi,\theta)(r,\vartheta)=(r,\vartheta)$.
The rotational symmetry of the geometry implies that the curves of constant $\theta$ must be geodesics; and our chart is already such that the curves $s \mapsto (s,\theta)$ are unit-speed. Thus (when restricted to $B_\pi$) the exponential map is exactly our chart - i.e. we have been using exponential coordinates. Since charts are diffeomorphisms onto their images this means that there are no conjugate points in the image of the chart.
It remains to check the antipodal point $q$ of $p$. To see that it is conjugate, let $v(t) = (\pi, t)$ be the parametrized boundary of $B_\pi \subset T_p S^2$. Every point on this curve is mapped to $q$ by the exponential map; so we have $$D_{v(0)} \exp(v'(0)) = \frac{d}{dt} \bigg|_{t=0} \exp(v(t)) = \frac{d}{dt} \bigg|_{t=0} q = 0.$$
