local diffeomorphism of exponential map for the circle Let $S^1$ be the Riemannian manifold with round metric. Pick a point $p \in S^1$ we have $\exp_p:T_pS^1 \to S^1$ globally defined. We know that $\exp_p(\pi) = q$ which is the antipodal point to $p$, which is conjugate to $p$, hence $\exp_p$ is critical at point $\pi \in T_pS^1$ correct?
We also know that $\exp_p$ is diffeomorphism on $(\frac{\pi}{2},\frac{3\pi}{2})$,which means $\exp_p$ is non singular at $\pi$. There is some contradiction.
What's wrong with my understanding?
 A: Your error is saying that $q = \exp_p(v)$ is a conjugate point of $p$ if and only if $\mathrm{d}\exp_p(v)$ is not injective. The right definition is the following: $p$ and $q$ are conjugate along the geodesic $\gamma$ if there exists a non-zero Jacobi field $J$ along $\gamma$ with $J(p)=J(q)=0$.
Here, you can show that such a Jacobi field does not exists on $S^1$.
Moreover, what makes the exponential map $\exp_p$ fail to be a diffeomorphism isn't any differential problem: it is a matter of injectivity, as $\exp_p(\pi) = \exp_p(-\pi) = -p$. It is easy to show that the exponential map $\exp_p$ is always a local diffeomorphism: if one writes $p = (\cos \theta, \sin \theta)$, then $\exp_p(t) = (\cos(\theta+t),\sin(\theta+t))$, and the differential is always non-zero.
A: You are working with $S^1$, the one-dimensional sphere and thus $-p$ is NOT conjugate to $p$: Since the curvature of $S^1$ is identically zero, the Jacobi equation has only linear solution: there is no non-trivial Jacobi field with $J(p) = J(-p) = 0$.
It is true that $\exp_p$ is a local diffeomorphism and there is no contradiction.
