The following is Exercise 10.1 in Riemannian Geometry by M. do Carmo.
(Klingenberg's Lemma). Let $M$ be a complete Riemannian manifold with sectional curvature $K<K_0$, where $K_0$ is a positive constant. Let $p,q\in M$ and let $\gamma_0$ and $\gamma_1$ be two distinct geodesics joining $p$ to $q$ with $\ell(\gamma_0)<\ell(\gamma_1)$. Assume that $\gamma_0$ is homotopic to $\gamma_1$, that is, there exists a continuous family of curves $\alpha_t$, $t\in[0,1]$ such that $\alpha_0=\gamma_0$ and $\alpha_1=\gamma_1$. Prove that there exists $t_0\in(0,1]$ such that $$\ell(\gamma_0)+\ell(\alpha_{t_0})\geq\frac{2\pi}{\sqrt{K_0}}.$$
The hint goes:
Hint: Assume $\ell(\gamma_0)<\pi/\sqrt{K_0}$ (otherwise, we have nothing to prove). From Ranch's Theorem, $\exp_p:TpM\to M$ has no critical point in the open ball $B$ of radius $\pi/\sqrt{K_0}$, centered at $p$. For $t$ small, it is possible to lift the curve at to the tangent space $T_pM$, i.e., there exists a curve $\widetilde{\alpha}_t$ in $T_pM$, joining $\exp_p^{-1}(0)=0$ to $\exp_p^{-1}(q)=\widetilde{q}$, such that $\exp_p\circ\widetilde{\alpha}_t=\alpha_t$. It is clear that it is not possible to do the same for every $t\in[0,1]$, since $\gamma_1$ cannot be lifted keeping the endpoints fixed.
We conclude that for all $\varepsilon>0$ there exists a $t(\varepsilon)$ such that $\alpha_{t(\varepsilon)}$ can be lifted to $\tilde{\alpha}_{t(\varepsilon)}$ and $\tilde{\alpha}_{t(\varepsilon)}$ contains points with distance
$<\varepsilon$ from the boundary $\partial B$ of $B$. In the contrary case, for some $\varepsilon>0$, all lifts $\tilde{\alpha}_t$ are at the distance $\geq\varepsilon$ from $\partial B$; the set of $t$'s for which it is possible to lift $\alpha_t$ will then be open and closed and $\alpha_1$ could be lifted, which is a contradiction. Therefore, for all $\varepsilon>0$, we have $$\ell(\gamma_0)+\ell(\alpha_{t(\varepsilon)})\geq\frac{2\pi}{\sqrt{K_0}}-\varepsilon.$$ Now choose a sequence $\{\varepsilon_n\}\to0$, and consider a convergent subsequence of $\{t(\varepsilon_n)\}\to t_0$. Then there exists a curve $\alpha_{t_0}$ with $$\ell(\gamma_0)+\ell(\alpha_{t_0})\geq\frac{2\pi}{\sqrt{K_0}}.$$
Why do such liftings exist? We only know that $\exp_p$ is nonsingular on $B(0,R):=\{v\in T_pM:|v|<R\}$, not that $\exp_p|_{B(0,R)}$ is a covering map or anything. Local diffeomorphisms can behave badly when it comes to lifting curves! So here is my question:
Let $(M,g)$ be a complete Riemannian manifold and $p\in M$. Suppose $\exp_p$ is nonsingular everywhere on $B(0,R)\subset T_pM$. Does any curve on $M$ starting from $p$ with length $<R$ lift to a curve on $T_pM$ starting at $0$? What about homotopies of such curves? In particular, why does the hint works?
Another question:
While in do Carmo's book this result is called Klingenberg's lemma, I cannot find it in any other resource. When and in which paper did Klingenberg prove this?
Best Answer
Regarding the question:
The answer to this in general is negative: Some curves do lift but some do not. However, the statement in your question is not what do Carmo is claiming. He only claims that if $\gamma: [0,a]\to M$ is a geodesic with $\gamma(0)=p$ then for all sufficiently small $T>0$ the restriction $\gamma|_{[0,T]}$ lifts to $T_pM$ via the exponential map to a radial line segment. This statement is obvious since $\exp_p$ is a local diffeomorphism at $0\in T_pM$.
Edit. I indeed misread the question. Here is the correct answer. One needs to assume that the curve $\alpha_0$ lifts. The map $\exp_p$ is a local diffeomorphism on $B(0, \pi/\sqrt{K_0})$. The same argument as in the proof of the covering homotopy theorem yields:
Lemma. Suppose that $f: X\to Y$ is a local homeomorphism between manifolds, $H:[0,1]^2\to Y$ is a continuous map and the curve $H(s,0), s\in [0,1]$ lifts to a curve $\tilde\alpha_0(s)=\tilde{H}(s,0)$ in $X$ via $f$. Then there exists $\epsilon>0$ such that the restriction of $H$ to $[0,1]\times [0,\epsilon]$ lifts to a map $\tilde{H}:[0,1]\times [0,\epsilon]\to X$ via $f$, whose restriction to $[0,1]\times\{0\}$ is the curve $\tilde\alpha_0$.
This lemma yields the claim made by do Carmo.