[proofplan] We use the standard general-topology criterion for covering maps: a surjective local diffeomorphism $f : M \to N$ between connected smooth manifolds is a covering map if and only if it has the homotopy lifting property for paths — equivalently, every smooth curve $\gamma : [0, 1] \to N$ admits a lift $\tilde\gamma : [0, 1] \to M$ starting at any prescribed point in $f^{-1}(\gamma(0))$. Local lifting exists in a small initial interval because $f$ is a local diffeomorphism. We argue that the supremum of times up to which a lift extends is open and closed in $[0, 1]$, the closedness coming from the metric estimate $|d f_p(v)|_h \geq |v|_g$ — which makes lifts uniformly Lipschitz, hence Cauchy at the supremum, hence convergent to a limit point in the complete manifold $M$ from which the lift extends past the supremum. Surjectivity is given. The path-lifting criterion then upgrades to the covering-map property via the standard fact for local diffeomorphisms with unique path lifting that any evenly covered neighbourhood structure follows. [/proofplan]

[step:Reduce to verifying the path-lifting property for smooth curves]We verify $f : M \to N$ is a covering map via the following standard criterion from covering-space theory for surjective local diffeomorphisms between connected smooth manifolds: **Criterion.** A surjective local diffeomorphism $f : M \to N$ between smooth manifolds with $N$ connected is a covering map if and only if it has the path-lifting property: every smooth path $\gamma : [0, 1] \to N$ has, for every $\tilde p \in f^{-1}(\gamma(0))$, a unique smooth lift $\tilde\gamma : [0, 1] \to M$ with $\tilde\gamma(0) = \tilde p$ and $f \circ \tilde\gamma = \gamma$. This is a standard result in covering-space theory (see, e.g., Lee, Smooth Manifolds, Chapter 4). Granted this criterion, our task reduces to producing such a lift for every smooth $\gamma : [0, 1] \to N$ and every $p_0 \in f^{-1}(\gamma(0))$. Surjectivity of $f$ is given by hypothesis, so $f^{-1}(\gamma(0))$ is non-empty.[/step]

[guided]The criterion is folklore-level: for a surjective local diffeomorphism between connected manifolds, evenly-covered neighbourhood structure is equivalent to the unique path-lifting property. The forward direction is direct (cover $\gamma$ by evenly covered neighbourhoods, lift segment by segment, glue). The converse direction takes a smooth bump path through any open neighbourhood and lifts it to produce the disjoint sheets above an evenly-covered neighbourhood. We mention the criterion explicitly and reduce the problem to path-lifting, which is the analytic content of the proof: whether the lift can be extended past every accumulation point. The completeness of $M$ and the metric inequality $|df_p(v)|_h \geq |v|_g$ enter precisely through this extension question.[/guided]

[step:Establish local lifting via the local-diffeomorphism property] Fix a smooth curve $\gamma : [0, 1] \to N$ and $p_0 \in f^{-1}(\gamma(0))$. Since $f$ is a local diffeomorphism at $p_0$, there exists an open neighbourhood $U \subseteq M$ of $p_0$ such that $f(U) \subseteq N$ is open and $f|_U : U \to f(U)$ is a diffeomorphism. Set $V := f(U)$. Continuity of $\gamma$ at $0$ gives $\varepsilon_0 > 0$ such that $\gamma([0, \varepsilon_0]) \subseteq V$. Define the local lift on $[0, \varepsilon_0]$: \begin{align*} \tilde\gamma : [0, \varepsilon_0] &\to U \subseteq M, \\ t &\mapsto (f|_U)^{-1}(\gamma(t)). \end{align*} Then $\tilde\gamma$ is smooth, $\tilde\gamma(0) = p_0$, and $f \circ \tilde\gamma = \gamma$ on $[0, \varepsilon_0]$. Define \begin{align*} T := \sup\{ s \in [0, 1] : \text{there exists a smooth lift } \tilde\gamma : [0, s] \to M \text{ of } \gamma|_{[0, s]} \text{ with } \tilde\gamma(0) = p_0 \}. \end{align*} By the local lift, $T \geq \varepsilon_0 > 0$. We claim $T = 1$ and the lift extends to $t = T = 1$. We prove this by showing the set of $s$ for which a lift exists is both open and closed in $(0, 1]$. [/step]

[step:Show the lift extends past every interior point: openness] Let $s \in [0, 1)$ be a point at which a lift $\tilde\gamma : [0, s] \to M$ exists. We show the lift extends to $[0, s + \varepsilon]$ for some $\varepsilon > 0$. Apply the local-diffeomorphism property at $\tilde\gamma(s) \in M$: there exists an open neighbourhood $U' \subseteq M$ of $\tilde\gamma(s)$ such that $f|_{U'} : U' \to f(U')$ is a diffeomorphism with $V' := f(U') \subseteq N$ open. By continuity of $\gamma$ at $s$, there exists $\varepsilon \in (0, 1 - s]$ such that $\gamma([s, s + \varepsilon]) \subseteq V'$. Define the extended lift on $[s, s + \varepsilon]$: \begin{align*} t \mapsto (f|_{U'})^{-1}(\gamma(t)) \in U'. \end{align*} At $t = s$, the right-hand side is $(f|_{U'})^{-1}(\gamma(s)) = (f|_{U'})^{-1}(f(\tilde\gamma(s))) = \tilde\gamma(s)$, so the new local definition agrees with the existing $\tilde\gamma$ at $s$. Concatenating the two pieces produces a smooth lift on $[0, s + \varepsilon]$. Hence the set of $s \in (0, 1]$ at which a lift exists is open in $(0, 1]$. [/step]

[step:Show the lift extends to its supremum: closedness via the metric inequality and completeness]Let $T \in (0, 1]$ be the supremum from Step 2; suppose $T > 0$. We show the lift extends to $[0, T]$. Pick a sequence $s_n \uparrow T$ with $s_n \in [0, T)$ and lifts $\tilde\gamma : [0, s_n] \to M$. Each lift agrees with the others on the common domain by uniqueness of lifts (which follows from the local-diffeomorphism property: any two lifts of $\gamma|_{[0, s_n]}$ starting at $p_0$ are equal, since the set where they agree is non-empty, open, and closed in $[0, s_n]$). So we have a single coherent lift $\tilde\gamma : [0, T) \to M$. We show $(\tilde\gamma(s_n))_{n \geq 1}$ is Cauchy in $(M, d_g)$, where $d_g$ is the Riemannian distance on $(M, g)$. Bound the velocity of the lift. For $t \in [0, T)$ at which $\tilde\gamma$ is differentiable: \begin{align*} df_{\tilde\gamma(t)}(\dot{\tilde\gamma}(t)) = \dot\gamma(t), \end{align*} by differentiating $f \circ \tilde\gamma = \gamma$. By the hypothesis $|df_p(v)|_h \geq |v|_g$ for all $v \in T_p M$ applied with $p = \tilde\gamma(t)$ and $v = \dot{\tilde\gamma}(t)$: \begin{align*} |\dot{\tilde\gamma}(t)|_g \leq |df_{\tilde\gamma(t)}(\dot{\tilde\gamma}(t))|_h = |\dot\gamma(t)|_h. \end{align*} Hence $|\dot{\tilde\gamma}(t)|_g \leq |\dot\gamma(t)|_h$ pointwise on $[0, T)$. Compute the $g$-length of the lift on the interval $[s_n, s_m]$ (with $s_n < s_m$, $s_m < T$): \begin{align*} \ell_g(\tilde\gamma|_{[s_n, s_m]}) = \int_{s_n}^{s_m} |\dot{\tilde\gamma}(t)|_g \, d\mathcal{L}^1(t) \leq \int_{s_n}^{s_m} |\dot\gamma(t)|_h \, d\mathcal{L}^1(t) = \ell_h(\gamma|_{[s_n, s_m]}). \end{align*} Therefore \begin{align*} d_g(\tilde\gamma(s_n), \tilde\gamma(s_m)) \leq \ell_g(\tilde\gamma|_{[s_n, s_m]}) \leq \ell_h(\gamma|_{[s_n, s_m]}) = \int_{s_n}^{s_m} |\dot\gamma(t)|_h \, d\mathcal{L}^1(t). \end{align*} The integrand $t \mapsto |\dot\gamma(t)|_h$ is continuous on the compact interval $[0, 1]$, hence bounded by some $L > 0$: \begin{align*} d_g(\tilde\gamma(s_n), \tilde\gamma(s_m)) \leq L \cdot |s_m - s_n|. \end{align*} Since $(s_n)$ converges to $T$, it is Cauchy in $\mathbb{R}$, and the inequality above shows $(\tilde\gamma(s_n))$ is Cauchy in $(M, d_g)$. By completeness of $(M, g)$ (which by [Hopf–Rinow Theorem](/theorems/2726) is equivalent to metric completeness of $(M, d_g)$), the Cauchy sequence converges: \begin{align*} \tilde\gamma(s_n) \to q^* \in M \quad \text{as } n \to \infty. \end{align*} The limit $q^*$ does not depend on the choice of sequence $(s_n) \uparrow T$: any two sequences $(s_n), (s_n') \uparrow T$ can be merged into a single sequence converging to $T$, and the merged image-sequence is Cauchy with the same limit; both sub-sequences inherit the limit, forcing equality. Continuity of $f$ gives $f(q^*) = \lim f(\tilde\gamma(s_n)) = \lim \gamma(s_n) = \gamma(T)$, so $q^* \in f^{-1}(\gamma(T))$. Define $\tilde\gamma(T) := q^*$. By the openness step applied at $\tilde\gamma(T)$, the lift extends to $[0, T + \varepsilon]$ for some $\varepsilon > 0$ if $T < 1$. If $T = 1$, the lift is defined on $[0, 1]$, completing the construction. Combined with the openness step, $T = 1$: if $T < 1$, the openness step contradicts the definition of $T$ as the supremum. Hence the lift exists on $[0, 1]$.[/step]

[guided]This is the heart of the proof. The danger we must rule out: as the parameter $t$ approaches $T$ from below, the lift $\tilde\gamma(t) \in M$ could escape to infinity, with no limit point in $M$ from which to continue. The local-diffeomorphism property alone does not prevent this — for example, the inclusion $(0, 1) \hookrightarrow \mathbb{R}$ is a local diffeomorphism, but a curve in $\mathbb{R}$ approaching $0$ has its lift escape from $(0, 1)$. We need a non-shrinking property of $f$ together with completeness of $M$ to close the gap. Let $T \in (0, 1]$ be the supremum from Step 2. We show the lift extends to $[0, T]$. **Coherence of the partial lifts.** First we assemble a single lift $\tilde\gamma : [0, T) \to M$. Pick $s_n \uparrow T$ with $s_n \in [0, T)$ and corresponding lifts $\tilde\gamma : [0, s_n] \to M$. Why are these compatible on overlaps? If two lifts $\tilde\gamma_1, \tilde\gamma_2 : [0, s] \to M$ both start at $p_0$ and both project to $\gamma$, the set $\{ t : \tilde\gamma_1(t) = \tilde\gamma_2(t)\}$ is non-empty (contains $0$), closed (by continuity), and open (by the local-diffeomorphism property: in a local-diffeomorphism neighbourhood of $\tilde\gamma_1(t_0)$, the inverse of $f$ is unique, so the two lifts must agree on a neighbourhood of $t_0$). A non-empty clopen subset of the connected interval $[0, s]$ is the whole interval. Hence the lifts on $[0, s_n]$ for varying $n$ glue into a single $\tilde\gamma : [0, T) \to M$. **Velocity of the lift.** The crucial estimate. Differentiating the relation $f \circ \tilde\gamma = \gamma$ at any $t \in [0, T)$ where the lift is differentiable gives \begin{align*} df_{\tilde\gamma(t)}(\dot{\tilde\gamma}(t)) = \dot\gamma(t). \end{align*} Now apply the hypothesis $|df_p(v)|_h \geq |v|_g$ with $p = \tilde\gamma(t)$ and $v = \dot{\tilde\gamma}(t)$: \begin{align*} |\dot{\tilde\gamma}(t)|_g \leq |df_{\tilde\gamma(t)}(\dot{\tilde\gamma}(t))|_h = |\dot\gamma(t)|_h. \end{align*} This is the entire reason the lift is well-behaved. The non-shrinking hypothesis forces the lift's $g$-speed to be bounded by the curve's $h$-speed: $f$ does not crush vectors, so the lift cannot move faster than the curve. **From velocity bound to length bound.** For $s_n < s_m < T$, integrate the velocity bound to bound the $g$-length of the lift on $[s_n, s_m]$ by the $h$-length of $\gamma$ on the same interval: \begin{align*} \ell_g(\tilde\gamma|_{[s_n, s_m]}) = \int_{s_n}^{s_m} |\dot{\tilde\gamma}(t)|_g \, d\mathcal{L}^1(t) \leq \int_{s_n}^{s_m} |\dot\gamma(t)|_h \, d\mathcal{L}^1(t) = \ell_h(\gamma|_{[s_n, s_m]}). \end{align*} The Riemannian distance is bounded by length of any connecting curve, so \begin{align*} d_g(\tilde\gamma(s_n), \tilde\gamma(s_m)) \leq \ell_g(\tilde\gamma|_{[s_n, s_m]}) \leq \ell_h(\gamma|_{[s_n, s_m]}) = \int_{s_n}^{s_m} |\dot\gamma(t)|_h \, d\mathcal{L}^1(t). \end{align*} **Cauchy condition.** Why does this make $(\tilde\gamma(s_n))$ Cauchy? The map $t \mapsto |\dot\gamma(t)|_h$ is continuous on the compact interval $[0, 1]$ (since $\gamma$ is smooth and $h$ is smooth), hence bounded by some $L > 0$. Therefore \begin{align*} d_g(\tilde\gamma(s_n), \tilde\gamma(s_m)) \leq L \cdot |s_m - s_n|. \end{align*} The right-hand side is just $L$ times the gap between two parameter values converging to $T$. Since $(s_n)$ is Cauchy in $\mathbb{R}$ (it converges to $T$), the image sequence $(\tilde\gamma(s_n))$ is Cauchy in $(M, d_g)$. **Convergence via completeness.** Now we cash in the completeness hypothesis. We verify the hypotheses of the [Hopf–Rinow Theorem](/theorems/2726): $(M, g)$ is given as a complete Riemannian manifold by hypothesis, which by Hopf–Rinow is equivalent to metric completeness of $(M, d_g)$. Hence every Cauchy sequence in $(M, d_g)$ converges: \begin{align*} \tilde\gamma(s_n) \to q^* \in M \quad \text{as } n \to \infty. \end{align*} **Independence of the sequence.** The limit $q^*$ does not depend on the choice of approximating sequence $(s_n) \uparrow T$. Given two such sequences $(s_n), (s_n') \uparrow T$, interleave them into a single sequence $(t_k) \uparrow T$. The same Cauchy argument applied to $(t_k)$ produces a Cauchy image-sequence with a unique limit. Both $(\tilde\gamma(s_n))$ and $(\tilde\gamma(s_n'))$ are subsequences of $(\tilde\gamma(t_k))$, so both inherit this same limit. Hence $q^*$ depends only on $T$. **Identifying $f(q^*)$.** Continuity of $f$ and continuity of $\gamma$ give \begin{align*} f(q^*) = \lim_{n \to \infty} f(\tilde\gamma(s_n)) = \lim_{n \to \infty} \gamma(s_n) = \gamma(T), \end{align*} so $q^* \in f^{-1}(\gamma(T))$. Define $\tilde\gamma(T) := q^*$. The smoothness of the extended lift at $T$ follows because in a local-diffeomorphism neighbourhood of $q^*$, the lift coincides with $(f|_{U'})^{-1} \circ \gamma$ for some $U'$, which is smooth. **Closing the induction.** If $T < 1$, the openness step (Step 3) applied at $\tilde\gamma(T) = q^*$ extends the lift to $[0, T + \varepsilon]$ for some $\varepsilon > 0$, contradicting $T = \sup$. Hence $T = 1$, and the construction above provides $\tilde\gamma(1)$. The lift exists on the full interval $[0, 1]$.[/guided]

[step:Conclude $f$ is a covering map]By the previous steps, every smooth curve $\gamma : [0, 1] \to N$ and every $p_0 \in f^{-1}(\gamma(0))$ admit a smooth lift $\tilde\gamma : [0, 1] \to M$ starting at $p_0$. By the criterion of Step 1, $f$ is a covering map. Connectedness of $M$ and $N$: $N$ is given (chapter standing hypothesis on Riemannian manifolds). $M$ is also assumed to be a Riemannian manifold (connected), so the argument applies in full generality. Surjectivity is part of the hypothesis.[/step]

[guided]The whole proof boils down to: smooth bumps in $N$ lift to $M$, and the lifts are uniformly bounded by the original curve's $h$-length, so they cannot escape $M$. Completeness of $M$ closes the gap at every limit point. The hypothesis $|df_p(v)|_h \geq |v|_g$ is the analytic content. It is a "non-shrinking" condition: $f$ does not crush vectors. Without it, the lift's velocity could blow up at limit points and the lift could escape compact sets of $M$ as we approach the supremum, even with $M$ complete. This same theorem applies to local isometries (where $|df_p(v)|_h = |v|_g$) and gives the Hadamard–Cartan covering result for the exponential map, where $f = \exp_p$ pulled back to its image.[/guided]