[proofplan] We prove $\operatorname{diam}(M, g) \leq \pi r$ by contradiction: if there exist $p, q \in M$ with $d(p, q) > \pi r$, completeness gives a unit-speed minimal geodesic $\gamma : [0, L] \to M$ from $p$ to $q$ with $L = d(p, q) > \pi r$. Length-minimality forces the index form $I(V, V) \geq 0$ for every endpoint-vanishing normal field $V$. We construct $n - 1$ test fields $V_i(t) := \sin(\pi t / L) E_i(t)$, where $(E_1, \dots, E_{n-1})$ is a parallel orthonormal frame for $\dot\gamma^\perp$ along $\gamma$. Direct computation gives $I(V_i, V_i) = \int_0^L \sin^2(\pi t / L) \big[ \pi^2/L^2 - R(E_i, \dot\gamma, E_i, \dot\gamma) \big] d\mathcal{L}^1(t)$ (after one integration by parts to convert the $|V_i'|^2$ kernel from $\cos^2$ to $\sin^2$). Summing over $i$ contracts the curvature term to the Ricci pairing $\sum_i R(E_i, \dot\gamma, E_i, \dot\gamma) = \operatorname{Ric}(\dot\gamma, \dot\gamma)$ in the chapter convention, and the Ricci hypothesis $\operatorname{Ric} \geq (n-1)/r^2 g$ combined with $L > \pi r$ makes $\sum_i I(V_i, V_i) < 0$, contradicting $I(V_i, V_i) \geq 0$. Hence $\operatorname{diam}(M, g) \leq \pi r$. Compactness follows from completeness plus bounded diameter via [Hopf–Rinow Theorem](/theorems/2726). Pulling back to the universal cover $\widetilde M$ — which inherits the same Ricci-curvature bound and is complete — gives $\widetilde M$ also compact, so the discrete fibre $\pi_1(M) = \pi^{-1}(p_0)$ inside the compact $\widetilde M$ is finite. [/proofplan]

[step:Reduce to a unit-speed minimal geodesic of length $L > \pi r$] Suppose for contradiction that $\operatorname{diam}(M, g) > \pi r$. Then there exist $p, q \in M$ with \begin{align*} L := d(p, q) > \pi r. \end{align*} By completeness of $(M, g)$ and [Minimal Geodesics From a Complete Point](/theorems/2724) applied at $p$, there exists a minimal geodesic from $p$ to $q$ of length $L$. By [Geodesics Have Constant Speed](/theorems/2709) and reparametrisation, we may assume this minimal geodesic is unit-speed: \begin{align*} \gamma : [0, L] \to M, \quad \gamma(0) = p, \quad \gamma(L) = q, \quad |\dot\gamma|_g \equiv 1, \quad \ell(\gamma) = L. \end{align*} [/step]

[step:Construct an orthonormal parallel frame along $\gamma$ orthogonal to $\dot\gamma$] Pick an orthonormal basis $(e_1, \ldots, e_{n-1})$ of $\dot\gamma(0)^\perp \subseteq T_p M$. Define the parallel transports \begin{align*} E_i : [0, L] &\to TM \quad \text{(section of $\gamma^* TM$)}, \\ E_i(t) &:= P_\gamma^{0, t}(e_i), \quad i = 1, \ldots, n - 1, \end{align*} where $P_\gamma^{0, t}$ is parallel transport along $\gamma$ from $0$ to $t$. Each $E_i$ is smooth on $[0, L]$, satisfies $\nabla_t E_i = 0$, and the orthonormality propagates by parallel transport (a linear isometry of tangent spaces): \begin{align*} g(E_i(t), E_j(t)) = \delta_{ij}, \qquad g(E_i(t), \dot\gamma(t)) = 0, \end{align*} where the second identity uses that $\dot\gamma$ is also parallel along $\gamma$ (geodesic equation $\nabla_t \dot\gamma = 0$). Hence $(E_1(t), \ldots, E_{n-1}(t))$ is an orthonormal basis of $\dot\gamma(t)^\perp$ for every $t \in [0, L]$. [/step]

[step:Define the test fields $V_i(t) := \sin(\pi t / L) E_i(t)$ and compute their derivatives] Define \begin{align*} V_i : [0, L] &\to TM \quad \text{(section of $\gamma^* TM$)}, \\ V_i(t) &:= \sin\!\left(\frac{\pi t}{L}\right) E_i(t), \quad i = 1, \ldots, n - 1. \end{align*} Each $V_i$ is smooth, vanishes at endpoints (since $\sin(0) = \sin(\pi) = 0$), and is pointwise normal to $\dot\gamma$ (each $E_i \perp \dot\gamma$). Using the Leibniz rule for $\nabla_t$ together with $\nabla_t E_i = 0$: \begin{align*} V_i'(t) := \nabla_t V_i(t) = \frac{\pi}{L} \cos\!\left(\frac{\pi t}{L}\right) E_i(t). \end{align*} Differentiating once more: \begin{align*} V_i''(t) := \nabla_t \nabla_t V_i(t) = -\frac{\pi^2}{L^2} \sin\!\left(\frac{\pi t}{L}\right) E_i(t) = -\frac{\pi^2}{L^2} V_i(t). \end{align*} [/step]

[step:Apply the second variation of length to obtain $I(V_i, V_i) \geq 0$] Since $\gamma$ minimises length among curves from $p$ to $q$, for every smooth endpoint-fixing variation $H : [0, L] \times (-\varepsilon, \varepsilon) \to M$ of $\gamma$, \begin{align*} \frac{d^2}{ds^2} \ell(H(\cdot, s))\Big|_{s = 0} \geq 0. \end{align*} Realise the variation via $H_i(t, s) := \exp_{\gamma(t)}(s V_i(t))$ — the exponential map at $\gamma(t)$ applied to the scaled normal field. This variation fixes the endpoints because $V_i(0) = V_i(L) = 0$, has variation field $\partial_s H_i(t, 0) = V_i(t)$, and is smooth on the rectangle $[0, L] \times (-\varepsilon_0, \varepsilon_0)$ for some $\varepsilon_0 > 0$ (taking $\varepsilon_0$ small enough that $\exp_{\gamma(t)}$ is defined on $\{s V_i(t) : |s| < \varepsilon_0\}$ uniformly in $t$, which holds because the normal field $V_i$ is bounded on the compact $[0, L]$ and $\exp$ is smooth on a uniform neighbourhood of the zero section over $\gamma([0, L])$). By Part 2 of the [Second Variation Formula](/theorems/2729), since $V_i \perp \dot\gamma$ everywhere and $V_i$ vanishes at the endpoints: \begin{align*} \frac{d^2}{ds^2} \ell(H_i(\cdot, s))\Big|_{s = 0} = I(V_i, V_i) := \int_0^L \big[\, |V_i'(t)|_g^2 - R(V_i, \dot\gamma, V_i, \dot\gamma)(t) \,\big] \, d\mathcal{L}^1(t). \end{align*} The boundary term in the second variation formula vanishes: at $t \in \{0, L\}$, the variation field $\partial_s H_i(t, s) = s \cdot d(\exp_{\gamma(t)})_0(V_i(t)) + O(s^2) = s V_i(t) + O(s^2)$ uniformly in $t$, so $\partial_s H_i(0, s) \equiv 0$ and $\partial_s H_i(L, s) \equiv 0$ (since $V_i(0) = V_i(L) = 0$); differentiating in $s$ at $s = 0$ gives $\nabla_s \partial_s H_i(t, 0) = 0$ at $t \in \{0, L\}$, killing the boundary term. Therefore \begin{align*} I(V_i, V_i) \geq 0 \quad \text{for each } i = 1, \ldots, n - 1. \end{align*} [/step]

[step:Compute $I(V_i, V_i)$ and convert the $|V_i'|^2$ integral via integration by parts]Substitute the expressions from Step 3: \begin{align*} |V_i'(t)|_g^2 &= \left| \frac{\pi}{L} \cos\!\left(\frac{\pi t}{L}\right) E_i(t) \right|_g^2 = \frac{\pi^2}{L^2} \cos^2\!\left(\frac{\pi t}{L}\right), \\ R(V_i, \dot\gamma, V_i, \dot\gamma)(t) &= \sin^2\!\left(\frac{\pi t}{L}\right) \cdot R(E_i, \dot\gamma, E_i, \dot\gamma)(t), \end{align*} where the second identity uses multilinearity of $R$ in its first and third slots. We apply integration by parts to express $\int_0^L |V_i'|_g^2 \, d\mathcal{L}^1(t)$ in terms of $V_i''$ and $V_i$. Using metric compatibility (the Leibniz rule for $g(V_i', V_i)$ along $\gamma$): \begin{align*} \frac{d}{dt} g(V_i', V_i) = g(V_i'', V_i) + g(V_i', V_i'), \end{align*} hence \begin{align*} \int_0^L g(V_i', V_i') \, d\mathcal{L}^1(t) = g(V_i', V_i)\Big|_0^L - \int_0^L g(V_i'', V_i) \, d\mathcal{L}^1(t). \end{align*} The boundary terms vanish since $V_i(0) = V_i(L) = 0$. Using $V_i'' = -(\pi^2 / L^2) V_i$ from Step 3: \begin{align*} g(V_i'', V_i)(t) = -\frac{\pi^2}{L^2} g(V_i, V_i)(t) = -\frac{\pi^2}{L^2} \sin^2\!\left(\frac{\pi t}{L}\right). \end{align*} Therefore \begin{align*} \int_0^L |V_i'|_g^2 \, d\mathcal{L}^1(t) = \frac{\pi^2}{L^2} \int_0^L \sin^2\!\left(\frac{\pi t}{L}\right) d\mathcal{L}^1(t). \end{align*} Combining, the index form becomes \begin{align*} I(V_i, V_i) = \int_0^L \sin^2\!\left(\frac{\pi t}{L}\right) \left[ \frac{\pi^2}{L^2} - R(E_i, \dot\gamma, E_i, \dot\gamma)(t) \right] d\mathcal{L}^1(t). \end{align*}[/step]

[guided]The goal is to evaluate $I(V_i, V_i)$ in a useful closed form. Plugging the formulas from Step 3 into the index form, the kinetic term $|V_i'|^2$ produces a $\cos^2(\pi t / L)$ kernel while the curvature term carries a $\sin^2(\pi t / L)$ kernel — different kernels, which is awkward for the comparison we want to make. The plan: integrate by parts to convert the $\cos^2$ kernel into $\sin^2$, so both terms sit over a common positive weight. Begin by substituting $V_i(t) = \sin(\pi t / L) E_i(t)$ and $V_i'(t) = (\pi/L) \cos(\pi t/L) E_i(t)$ from Step 3. Since $|E_i(t)|_g = 1$ (Step 2): \begin{align*} |V_i'(t)|_g^2 &= \left| \frac{\pi}{L} \cos\!\left(\frac{\pi t}{L}\right) E_i(t) \right|_g^2 = \frac{\pi^2}{L^2} \cos^2\!\left(\frac{\pi t}{L}\right), \\ R(V_i, \dot\gamma, V_i, \dot\gamma)(t) &= \sin^2\!\left(\frac{\pi t}{L}\right) \cdot R(E_i, \dot\gamma, E_i, \dot\gamma)(t), \end{align*} where the second identity uses that $R$ is multilinear in its first and third slots, so the scalar factor $\sin(\pi t / L)$ pulls out twice — once from each $V_i$ slot — yielding $\sin^2(\pi t / L)$. Now we integrate by parts. The motivation: $V_i$ satisfies the eigenvalue equation $V_i'' = -(\pi^2/L^2) V_i$ along the parallel frame (Step 3), so $|V_i'|^2$ and $|V_i|^2$ are not independent — a Leibniz rule should let us trade one for the other. Concretely, by metric compatibility of $\nabla$ along $\gamma$: \begin{align*} \frac{d}{dt} g(V_i', V_i) = g(V_i'', V_i) + g(V_i', V_i'). \end{align*} Integrating from $0$ to $L$: \begin{align*} \int_0^L g(V_i', V_i') \, d\mathcal{L}^1(t) = g(V_i', V_i)\Big|_0^L - \int_0^L g(V_i'', V_i) \, d\mathcal{L}^1(t). \end{align*} Why does this help? The boundary terms $g(V_i', V_i)|_0^L$ vanish because $V_i(0) = V_i(L) = 0$ (the endpoint vanishing was built into the test field via $\sin(0) = \sin(\pi) = 0$). And the remaining integrand $g(V_i'', V_i)$ is computable: substituting $V_i'' = -(\pi^2/L^2) V_i$ from Step 3, \begin{align*} g(V_i'', V_i)(t) = -\frac{\pi^2}{L^2} g(V_i, V_i)(t) = -\frac{\pi^2}{L^2} \sin^2\!\left(\frac{\pi t}{L}\right). \end{align*} Therefore \begin{align*} \int_0^L |V_i'|_g^2 \, d\mathcal{L}^1(t) = \frac{\pi^2}{L^2} \int_0^L \sin^2\!\left(\frac{\pi t}{L}\right) d\mathcal{L}^1(t). \end{align*} The trigonometric identity $\cos^2 + \sin^2 = 1$ would have given the same answer through $\int_0^L \cos^2 = \int_0^L \sin^2 = L/2$, but the integration-by-parts derivation is the conceptually correct one: it works for any test field of the form $\phi(t) E_i(t)$ with $\phi(0) = \phi(L) = 0$, not just sines. Combining the two pieces — the converted $|V_i'|^2$ integral and the $R$ expression — the index form takes the unified form \begin{align*} I(V_i, V_i) = \int_0^L \sin^2\!\left(\frac{\pi t}{L}\right) \left[ \frac{\pi^2}{L^2} - R(E_i, \dot\gamma, E_i, \dot\gamma)(t) \right] d\mathcal{L}^1(t). \end{align*} Both terms now sit over the common positive kernel $\sin^2(\pi t / L)$, ready for the curvature comparison in the next step. A remark on the choice of test field. The eigenvalue $\pi^2/L^2$ is the smallest eigenvalue of the Dirichlet Laplacian $-d^2/dt^2$ on $[0, L]$ — exactly matching the constant-Ricci threshold $(n-1)/r^2$ when $L = \pi r$. The choice $V_i(t) = \sin(\pi t/L) E_i(t)$ is the optimal endpoint-vanishing test field: any other shape function $\phi$ would produce a larger Rayleigh quotient $\int \phi'^2 / \int \phi^2$ and correspondingly a weaker contradiction.[/guided]

[step:Sum over $i$ and convert the curvature trace to the Ricci tensor] Sum the $n - 1$ index forms: \begin{align*} \sum_{i = 1}^{n - 1} I(V_i, V_i) = \int_0^L \sin^2\!\left(\frac{\pi t}{L}\right) \left[ (n - 1) \cdot \frac{\pi^2}{L^2} - \sum_{i = 1}^{n - 1} R(E_i, \dot\gamma, E_i, \dot\gamma)(t) \right] d\mathcal{L}^1(t). \end{align*} We rewrite the curvature sum as the Ricci pairing. With the chapter sign convention $R = -\nabla \circ \nabla$ for the curvature operator and the corresponding Ricci tensor convention \begin{align*} \operatorname{Ric}(X, Y) := \sum_{k = 1}^{n} R(F_k, X, F_k, Y), \end{align*} in any orthonormal frame $(F_1, \ldots, F_n)$ of the tangent space — this is the convention paired with $R = -\nabla \circ \nabla$ such that $\operatorname{Ric} \geq c \cdot g$ encodes a positive-curvature lower bound consistent with classical Bonnet-Myers — we apply this with $X = \dot\gamma(t)$ and the orthonormal frame $E_1(t), \ldots, E_{n-1}(t), \dot\gamma(t)$ of $T_{\gamma(t)} M$: \begin{align*} \operatorname{Ric}(\dot\gamma, \dot\gamma)(t) = \sum_{k = 1}^{n} R(F_k, \dot\gamma, F_k, \dot\gamma)(t). \end{align*} The $k = n$ term, with $F_n = \dot\gamma$, vanishes by skew-symmetry of $R$ in the first pair: $R(\dot\gamma, \dot\gamma, \dot\gamma, \dot\gamma) = 0$ (see [Symmetries of the Riemann Curvature Tensor](/theorems/2704)). The remaining $k = 1, \ldots, n - 1$ terms have $F_k = E_k$, giving \begin{align*} \operatorname{Ric}(\dot\gamma, \dot\gamma)(t) = \sum_{i = 1}^{n - 1} R(E_i, \dot\gamma, E_i, \dot\gamma)(t). \end{align*} Substituting: \begin{align*} \sum_{i = 1}^{n - 1} I(V_i, V_i) = \int_0^L \sin^2\!\left(\frac{\pi t}{L}\right) \left[ (n - 1) \cdot \frac{\pi^2}{L^2} - \operatorname{Ric}(\dot\gamma, \dot\gamma)(t) \right] d\mathcal{L}^1(t). \end{align*} [/step]

[step:Apply the Ricci hypothesis to derive a contradiction]By hypothesis, $\operatorname{Ric}(g) \geq (n - 1)/r^2 \cdot g$, so for the unit vector $\dot\gamma(t)$: \begin{align*} \operatorname{Ric}(\dot\gamma, \dot\gamma)(t) \geq \frac{n - 1}{r^2}. \end{align*} Therefore the integrand in the previous step satisfies \begin{align*} (n - 1) \cdot \frac{\pi^2}{L^2} - \operatorname{Ric}(\dot\gamma, \dot\gamma)(t) \leq (n - 1)\!\left[ \frac{\pi^2}{L^2} - \frac{1}{r^2} \right]. \end{align*} Since $L > \pi r$, $L^2 > \pi^2 r^2$, hence $\pi^2 / L^2 < 1/r^2$, so $\pi^2/L^2 - 1/r^2 < 0$. The bracket is strictly negative. Multiplying by the non-negative kernel $\sin^2(\pi t / L) \geq 0$ (strictly positive on $(0, L)$) and integrating: \begin{align*} \sum_{i = 1}^{n - 1} I(V_i, V_i) \leq (n - 1)\!\left[ \frac{\pi^2}{L^2} - \frac{1}{r^2} \right] \cdot \int_0^L \sin^2\!\left(\frac{\pi t}{L}\right) d\mathcal{L}^1(t) = (n - 1)\!\left[ \frac{\pi^2}{L^2} - \frac{1}{r^2} \right] \cdot \frac{L}{2} < 0, \end{align*} using $\int_0^L \sin^2(\pi t / L) d\mathcal{L}^1(t) = L/2$. Hence $\sum_{i = 1}^{n - 1} I(V_i, V_i) < 0$, so at least one summand $I(V_{i_0}, V_{i_0}) < 0$. This contradicts $I(V_i, V_i) \geq 0$ for every $i$ from Step 4. The contradiction came from the assumption $\operatorname{diam}(M, g) > \pi r$. Therefore $\operatorname{diam}(M, g) \leq \pi r$.[/step]

[guided]The setup from Step 6 gave us the summed index form \begin{align*} \sum_{i = 1}^{n - 1} I(V_i, V_i) = \int_0^L \sin^2\!\left(\frac{\pi t}{L}\right) \left[ (n - 1) \cdot \frac{\pi^2}{L^2} - \operatorname{Ric}(\dot\gamma, \dot\gamma)(t) \right] d\mathcal{L}^1(t). \end{align*} The strategy now is clear: bound the bracketed integrand from above by something strictly negative using the Ricci hypothesis, then conclude the whole sum is negative — contradicting the lower bound $I(V_i, V_i) \geq 0$ from Step 4. The hypothesis says $\operatorname{Ric}(g) \geq (n - 1)/r^2 \cdot g$ as a tensor inequality, meaning $\operatorname{Ric}(X, X) \geq ((n-1)/r^2) g(X, X)$ for every tangent vector $X$. We apply this with $X = \dot\gamma(t)$, which is a unit vector by the unit-speed parametrisation (so $g(\dot\gamma, \dot\gamma) = 1$): \begin{align*} \operatorname{Ric}(\dot\gamma, \dot\gamma)(t) \geq \frac{n - 1}{r^2}. \end{align*} Substituting this lower bound into the bracketed expression gives an upper bound (the sign flips because $\operatorname{Ric}$ enters with a minus sign): \begin{align*} (n - 1) \cdot \frac{\pi^2}{L^2} - \operatorname{Ric}(\dot\gamma, \dot\gamma)(t) \leq (n - 1)\!\left[ \frac{\pi^2}{L^2} - \frac{1}{r^2} \right]. \end{align*} Now we use the contradiction hypothesis $L > \pi r$. Squaring (both sides positive) gives $L^2 > \pi^2 r^2$, hence $\pi^2/L^2 < 1/r^2$, hence $\pi^2/L^2 - 1/r^2 < 0$. The bracket is strictly negative, and the constant $(n - 1) > 0$ since $\dim M = n \geq 2$. So the upper bound on the integrand is strictly negative. This is the decisive inequality: \begin{align*} L > \pi r \;\implies\; \frac{\pi^2}{L^2} < \frac{1}{r^2} \;\implies\; (n - 1) \frac{\pi^2}{L^2} < \frac{n - 1}{r^2} \leq \operatorname{Ric}(\dot\gamma, \dot\gamma). \end{align*} The threshold $L = \pi r$ is exactly where the test-field "kinetic energy" $(n-1)\pi^2/L^2$ matches the curvature lower bound $(n-1)/r^2$. Past this threshold the curvature dominates, and the index form is forced negative. Multiply through by the kernel $\sin^2(\pi t / L) \geq 0$ (strictly positive on $(0, L)$, so the inequality is preserved without degenerating) and integrate over $[0, L]$: \begin{align*} \sum_{i = 1}^{n - 1} I(V_i, V_i) \leq (n - 1)\!\left[ \frac{\pi^2}{L^2} - \frac{1}{r^2} \right] \cdot \int_0^L \sin^2\!\left(\frac{\pi t}{L}\right) d\mathcal{L}^1(t) = (n - 1)\!\left[ \frac{\pi^2}{L^2} - \frac{1}{r^2} \right] \cdot \frac{L}{2} < 0, \end{align*} where we used $\int_0^L \sin^2(\pi t / L) \, d\mathcal{L}^1(t) = L/2$ (a direct computation from the half-angle identity $\sin^2 \theta = (1 - \cos 2\theta)/2$). The product on the right is strictly negative because each factor is positive except the bracket, which is strictly negative. So $\sum_{i = 1}^{n - 1} I(V_i, V_i) < 0$. Since the sum of $n - 1$ non-negative numbers cannot be negative, at least one summand $I(V_{i_0}, V_{i_0})$ must itself be negative — but this contradicts $I(V_i, V_i) \geq 0$ for every $i$ from Step 4 (where length-minimisation forced non-negativity of the second variation). The contradiction came from assuming $\operatorname{diam}(M, g) > \pi r$. Therefore $\operatorname{diam}(M, g) \leq \pi r$. A remark on why this particular test field works. In a constant-curvature space form $S^n_r$ of radius $r$ (where $\operatorname{Ric} = (n-1)/r^2 \cdot g$ exactly), the actual normal Jacobi fields along a unit-speed geodesic are precisely $\sin(t / r) E_i(t)$ in the orthonormal parallel frame. The Bonnet-Myers proof tests against the limit case: $\sin(\pi t / L)$ matches the sphere's Jacobi fields when $L = \pi r$ (the diameter of the sphere). At $L > \pi r$, the test field has a smaller Rayleigh quotient than the Ricci curvature, forcing a negative index form. The Ricci-trace identity $\operatorname{Ric}(X, X) = \sum_i R(E_i, X, E_i, X)$ from Step 6 is what converts the sectional-curvature index-form computation into a Ricci-curvature contradiction — exactly why a Ricci hypothesis (rather than a sectional hypothesis) suffices.[/guided]

[step:Deduce compactness of $M$] By the previous steps, $\operatorname{diam}(M, g) \leq \pi r$. Fix any $p_0 \in M$. For every $q \in M$, $d(p_0, q) \leq \operatorname{diam}(M) \leq \pi r$, so \begin{align*} M = \overline{B}_{d_g}(p_0, \pi r), \end{align*} the closed metric ball of radius $\pi r$ around $p_0$. By the [Hopf–Rinow Theorem](/theorems/2726) — specifically, the implication $(1) \implies (4)$ that complete connected Riemannian manifolds satisfy the Heine–Borel property (every closed bounded subset is compact) — the closed bounded set $M$ is compact. [/step]

[step:Deduce finiteness of $\pi_1(M)$ via the universal cover]Let $\pi : \widetilde M \to M$ be the universal cover with the pulled-back metric $\tilde g := \pi^* g$. Then $\pi$ is a smooth local isometry. We verify that Bonnet-Myers applies to $(\widetilde M, \tilde g)$: - $\widetilde M$ is connected by definition of the universal cover. - $\tilde g$ has the same Ricci-curvature lower bound: Ricci is a local invariant (a smooth $(0,2)$-tensor depending pointwise on the metric and its first two derivatives), and the local-isometry $\pi$ preserves these pointwise. For every $\tilde x \in \widetilde M$ and every $\tilde X \in T_{\tilde x} \widetilde M$, $\operatorname{Ric}(\tilde g)_{\tilde x}(\tilde X, \tilde X) = \operatorname{Ric}(g)_{\pi(\tilde x)}(d\pi_{\tilde x} \tilde X, d\pi_{\tilde x} \tilde X) \geq (n-1)/r^2 \cdot |d\pi_{\tilde x} \tilde X|_g^2 = (n-1)/r^2 \cdot |\tilde X|_{\tilde g}^2$. - $\widetilde M$ is complete. Geodesics of $\widetilde M$ project to geodesics of $M$ via $\pi$ (local isometries push geodesics to geodesics by uniqueness of the Levi-Civita connection of the pullback metric). If $\tilde\gamma : [0, b) \to \widetilde M$ is a maximal geodesic with $b < \infty$, then $\gamma := \pi \circ \tilde\gamma : [0, b) \to M$ is a geodesic of $(M, g)$. By completeness of $(M, g)$, $\gamma$ extends smoothly to $[0, b + \delta)$ for some $\delta > 0$. By the path-lifting property of the covering map $\pi$ (with the lifting starting at $\tilde\gamma(b - \varepsilon)$ for small $\varepsilon > 0$ and matching $\tilde\gamma$ on the overlap), $\tilde\gamma$ also extends to $[0, b + \delta)$, contradicting maximality. Hence $b = \infty$, and $\widetilde M$ is geodesically complete; equivalently metrically complete by the [Hopf–Rinow Theorem](/theorems/2726). By the diameter bound and compactness arguments (Steps 1–8 applied to $(\widetilde M, \tilde g)$): \begin{align*} \operatorname{diam}(\widetilde M, \tilde g) \leq \pi r, \quad \widetilde M \text{ is compact}. \end{align*} The deck-transformation group $\operatorname{Deck}(\widetilde M / M)$ is canonically isomorphic to $\pi_1(M)$ and acts freely and properly discontinuously on $\widetilde M$. The fibre over any $p_0 \in M$ satisfies \begin{align*} |\pi^{-1}(p_0)| = |\pi_1(M)|. \end{align*} The fibre $\pi^{-1}(p_0) \subseteq \widetilde M$ is a discrete subset of $\widetilde M$ (covering maps have discrete fibres). A discrete subset of a compact metric space is finite (any infinite set has an accumulation point in the compact space, contradicting discreteness). Hence \begin{align*} |\pi_1(M)| = |\pi^{-1}(p_0)| < \infty, \end{align*} so $\pi_1(M)$ is finite.[/step]

[guided]The strategy: lift to the universal cover, apply the diameter bound there, and read off finiteness of $\pi_1(M)$ from the size of a discrete fibre inside a compact space. The plan has three pieces — verify Bonnet-Myers applies to the universal cover, deduce its compactness, and translate compactness into finiteness of the fundamental group via the deck action. Let $\pi : \widetilde M \to M$ be the universal cover with the pulled-back metric $\tilde g := \pi^* g$. The map $\pi$ is then a smooth local isometry by construction of the pullback metric. We need to verify the three Bonnet-Myers hypotheses for $(\widetilde M, \tilde g)$. *Connectedness.* The universal cover is by construction path-connected (in fact simply connected — that is its defining property), so connectedness is immediate. *Ricci lower bound.* Why does the bound transfer? Ricci is a local invariant — a smooth $(0,2)$-tensor depending pointwise on the metric and its first two derivatives. A local isometry preserves the metric and hence all of its derivatives in coordinates, so it preserves Ricci pointwise. Concretely, for every $\tilde x \in \widetilde M$ and every $\tilde X \in T_{\tilde x} \widetilde M$: \begin{align*} \operatorname{Ric}(\tilde g)_{\tilde x}(\tilde X, \tilde X) = \operatorname{Ric}(g)_{\pi(\tilde x)}(d\pi_{\tilde x} \tilde X, d\pi_{\tilde x} \tilde X) \geq \frac{n-1}{r^2} \cdot |d\pi_{\tilde x} \tilde X|_g^2 = \frac{n-1}{r^2} \cdot |\tilde X|_{\tilde g}^2, \end{align*} where the first equality is the local-isometry pullback of $\operatorname{Ric}$, the inequality is the hypothesis on $(M, g)$, and the last equality is the definition $|\tilde X|_{\tilde g}^2 = |d\pi_{\tilde x} \tilde X|_g^2$ of the pullback metric. So $\operatorname{Ric}(\tilde g) \geq (n-1)/r^2 \cdot \tilde g$ on $\widetilde M$, with no loss in the constant. *Completeness.* Geodesics of $\widetilde M$ project to geodesics of $M$ via $\pi$ — local isometries push geodesics to geodesics by uniqueness of the Levi-Civita connection of the pullback metric. We argue completeness by contradiction. Suppose $\tilde\gamma : [0, b) \to \widetilde M$ is a maximal geodesic with $b < \infty$. Then $\gamma := \pi \circ \tilde\gamma : [0, b) \to M$ is a geodesic of $(M, g)$. By completeness of $(M, g)$, $\gamma$ extends smoothly to $[0, b + \delta)$ for some $\delta > 0$. The path-lifting property of the covering map $\pi$ — applied with the lifting starting at $\tilde\gamma(b - \varepsilon)$ for small $\varepsilon > 0$ and matching $\tilde\gamma$ on the overlap — produces a smooth extension of $\tilde\gamma$ to $[0, b + \delta)$, contradicting maximality. Hence $b = \infty$, so $\widetilde M$ is geodesically complete; equivalently metrically complete by the [Hopf–Rinow Theorem](/theorems/2726). With all three hypotheses verified, the diameter bound and compactness arguments from Steps 1–8 apply verbatim to $(\widetilde M, \tilde g)$, giving \begin{align*} \operatorname{diam}(\widetilde M, \tilde g) \leq \pi r, \quad \widetilde M \text{ is compact}. \end{align*} Now we extract finiteness of $\pi_1(M)$ from compactness of $\widetilde M$. The deck-transformation group $\operatorname{Deck}(\widetilde M / M)$ is canonically isomorphic to $\pi_1(M)$ and acts freely and properly discontinuously on $\widetilde M$. Why does this give us what we want? Because the deck action is *free* and *transitive on each fibre*: given any $p_0 \in M$ and any $\tilde p_0 \in \pi^{-1}(p_0)$, the orbit map $\operatorname{Deck}(\widetilde M / M) \to \pi^{-1}(p_0)$, $\sigma \mapsto \sigma(\tilde p_0)$, is a bijection. So \begin{align*} |\pi^{-1}(p_0)| = |\operatorname{Deck}(\widetilde M / M)| = |\pi_1(M)|. \end{align*} The fibre $\pi^{-1}(p_0) \subseteq \widetilde M$ is a discrete subset (covering maps have discrete fibres by definition — every point in the fibre has an evenly covered neighbourhood whose preimage is a disjoint union of sheets, each containing exactly one fibre point). A discrete subset $S$ of a compact metric space $\widetilde M$ must be finite. Why? Suppose $S$ is infinite. Pick a sequence of distinct points $\tilde x_k \in S$. By compactness of $\widetilde M$, this sequence has a convergent subsequence $\tilde x_{k_j} \to \tilde x_\infty$ in $\widetilde M$. But $\tilde x_\infty$ is then an accumulation point of $S$ in $\widetilde M$, contradicting discreteness (which says every point of $S$ has a neighbourhood in $\widetilde M$ containing no other point of $S$). Hence $S$ is finite. Applying this to $S = \pi^{-1}(p_0)$: \begin{align*} |\pi_1(M)| = |\pi^{-1}(p_0)| < \infty, \end{align*} so $\pi_1(M)$ is finite. This finishes the proof: a complete connected $n$-manifold with $\operatorname{Ric} \geq (n-1)/r^2 \cdot g$ has diameter $\leq \pi r$, is compact, and has finite fundamental group.[/guided]