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Embed radius-$L$ Anti-de Sitter spacetime as $X\cdot X=-L^2$ in signature with two timelike directions. Initial tangent $V_P$ satisfies $X_P\cdot V_P=0$ and $V_P\cdot V_P=-1$. The displayed curve remains on the hyperboloid and has unit timelike tangent; its ambient acceleration is normal, proving it is a geodesic. Every such curve reaches $-X_P$ after proper time $\pi L$. On the universal cover, this is a later antipodal event, not a closed causal loop. Global time advances by $2\pi$ over a complete ambient period.

In an orientable irreducible three-manifold, an incompressible cut is fundamental group-injective by the Loop theorem. Lift a three-manifold hierarchy to the universal cover. The terminal ball pieces are contractible and their gluing surfaces have contractible universal covers. The incidence graph is a tree, by the van Kampen theorem normal form for successive cuts. Finite subtrees are contractible by successively attaching a terminal piece along a contractible intersection. Every sphere map has compact image contained in a finite subtree, so all higher homotopy groups of the universal cover vanish.

The simply connected Bertotti-Robinson geometry is the product of the universal cover of two-dimensional anti-de Sitter spacetime and a two-sphere with equal radius $m>0$. In metric signature $(-,+,+,+)$,Its radial Penrose diagram is an infinite conformal strip of two-dimensional anti-de Sitter spacetime. The metric tensor $-{\lambda }^{2}d{t}^2+m^2{\lambda }^{-2}d{\lambda }^2+m^{2}d{\Omega }_2^2$ is a patch of it, with $\lambda =m(\sin T-\sin X)/\cos X$. Its Killing vector field ${\partial }_t$ extends across alternating degenerate Killing horizons $\lambda =0$. Both factors are geodesically complete, giving a maximal analytic extension of this patch without periodic time identification.

For a connected CW complex $X$ with contractible universal cover, let $\Gamma ={\pi }_1(X)$ and $R=\mathbb{Z}\Gamma$. Lift each cell once and transport its orientation by deck transformations. Its translates form an $R$-basis of the cellular chains of the cover. The augmentation sends every vertex to one, and contractibility makes the augmented chain complex an exact free resolution of the trivial module $\mathbb{Z}$. An $R$-linear cochain with trivial coefficient action is constant on each orbit of lifts, identifying ${\mathrm{Hom}}_R(C_{\ast }(\tilde{X}),\mathbb{Z})$ with the cellular cohomology complex of $X$. Consequently $H^i(X;\mathbb{Z})={\mathrm{Ext}}_R^i(\mathbb{Z},\mathbb{Z})$. The same argument with singular simplex lifts works for spaces with a contractible universal cover.

For a connected complete Riemannian manifold of dimension $n\ge 2$ and constant sectional curvature $k>0$, its complete simply connected universal cover is the round sphere of radius $k^{-1/2}$. This is the constant-curvature case of Cartan-Ambrose-Hicks theorem. Thus the manifold is a quotient by a finite group of isometries acting freely. In even dimension, the determinant embeds this group in $\{1,-1\}$: every determinant-one orthogonal transformation in odd ambient dimension has eigenvalue one and therefore a fixed point. The only possible nonidentity element is the antipodal involution, giving Real projective space. In odd dimensions, free cyclic scalar actions on complex unit spheres instead produce lens spaces.

Give a discrete group $G$ a contractible free CW complex $EG$. The quotient $BG=EG/G$ has universal cover $EG$, so it has fundamental group $G$ and no higher homotopy groups. It is thus an Eilenberg–MacLane space $K(G,1)$, even when $G$ is nonabelian. The cohomology of $BG$ with the appropriate local coefficients computes group cohomology.

The Complex projective plane is simply connected. If it were the total space of a covering of degree at least two, it would be the universal cover and have a nonidentity deck transformation. Such a transformation acts by $u\mapsto \pm u$ on $H^{\ast }({\mathbb{CP}}^2)=\mathbb{Z}[u]/(u^3)$, giving Lefschetz number 1 or 3. It therefore has a fixed point, impossible for a nonidentity deck transformation.

For a locally path-connected subspace $A\subseteq X$, a selected component of the inverse image in a universal cover covers the corresponding component $A_0$ of $A$. Its covering subgroup is $\ker ({\pi }_1(A,a_0)\to {\pi }_1(X,a_0))$; surjectivity onto all of $A$ additionally requires $A$ to be path connected.

If a connected complete Riemannian manifold has $\mathrm{Ric}\ge (n-1)\kappa g$ with $\kappa >0$, its universal cover with the lifted metric is complete and satisfies the same bound. The Bonnet-Myers theorem makes that cover compact. A covering fiber is closed and discrete and hence finite; the number of its elements equals the order of the fundamental group. Applying the diameter theorem only to the base would not prove this conclusion: the positive lower bound must also be used on the cover.

For a compact connected Riemannian manifold with nonnegative Ricci curvature, its finitely generated fundamental group acts freely and properly discontinuously on the complete universal cover. The preceding polynomial growth of properly discontinuous isometry groups bound applies. The degree-one Hurewicz theorem identifies first integral homology with the abelianization of the fundamental group. If its free rank is $b_1(M)$, this group maps onto ${\mathbb{Z}}^{b_1(M)}$. Choosing lifts of a basis gives at least a constant times $m^{b_1(M)}$ distinct elements in a word ball of radius a constant times $m$. The upper bound of order $m^{\dim M}$ forces the displayed inequality.

The universal cover of $S^1$ gives the first case. The Hopf fibration $S^1\to S^3\to S^2$ identifies ${\pi }_3(S^2)$ with ${\pi }_3(S^3)=\mathbb{Z}$. Starting from ${\pi }_4(S^3)=\mathbb{Z}/2$, the Freudenthal suspension theorem makes every subsequent suspension in these degrees an isomorphism. The stable range does not include the first step from ${\pi }_3(S^2)$ to ${\pi }_4(S^3)$, so the integer group at $n=2$ is not a contradiction.

A geodesic of the unit embedded hyperboloid lies in the plane spanned by its initial point and tangent. The constraint $X^2=1$ forces the displayed ambient acceleration. Timelike geodesics are $X=P\cos s+V\sin s$, null ones are $X=P+sV$, and spacelike ones are $X=P\cosh s+V\sinh s$. Timelike geodesics are closed on the hyperboloid but lift to complete nonclosed curves on its universal cover. In static coordinates, conserved $E={\cosh }^{2}rdt/ds$ gives $(dr/ds{)}^2=E^2/{\cosh }^{2}r-1$ and $\sinh r=\sqrt{E^2-1}\sin (s-s_0)$. Null affine parameters obey $\lambda =\pm \sinh r/E+\text{constant}$, proving completeness at both spatial ends.

Moving the basepoint around a loop acts on a higher homotopy group. Since ${\pi }_n(X)$ is an abelian group for $n\ge 2$, this action extends linearly to its integral group ring, making ${\pi }_n(X)$ a module. For a connected CW complex, its universal cover gives an equivalent description by deck transformations, together with change of basepoint upstairs. Left and right conventions are converted by inversion of the loop. The Hurewicz theorem respects this action whenever it identifies a higher homotopy group with the corresponding homology of the cover.

A complete hyperbolic three-manifold is a complete Riemannian manifold of dimension three with constant sectional curvature $-1$. Its universal cover is hyperbolic space ${\mathbb{H}}^3$ and its deck group acts freely and discretely by Riemannian isometries. In the orientable case it is a torsion-free Kleinian group.

All the special orthogonal groups are connected, and their second homotopy groups vanish. The rotation circle gives $SO(2)\cong S^1$. Conjugation by unit quaternions gives the twofold universal cover $S^3\to SO(3)$, so ${\pi }_{1}SO(3)=\mathbb{Z}/2$ and ${\pi }_{3}SO(3)=\mathbb{Z}$. The special orthogonal sphere fibration $SO(n-1)\to SO(n)\to S^{n-1}$ propagates the first two homotopy groups for $n\ge 4$. In dimension four its exact sequence has a zero map from the torsion group ${\pi }_4{S}^3=\mathbb{Z}/2$ to ${\pi }_{3}SO(3)=\mathbb{Z}$, giving ${\pi }_{3}SO(4)={\mathbb{Z}}^2$. For $SO(1)$ all positive groups vanish, and ${\pi }_{3}SO(2)=0$.

For a prime $p$, a classifying Eilenberg–MacLane space for $C_p$ is the infinite lens space $S^{\infty }/C_p$, using scalar multiplication by a primitive $p$th root on the complex-coordinate sphere. Its contractible universal cover is $S^{\infty }$. The standard two-periodic free resolution has differentials $g-1$ and $1+g+\cdots +g^{p-1}$, both zero after applying cochains with trivial ${\mathbb{F}}_p$ coefficients. Thus every cohomology degree has dimension one. If $p=2$, the ring is ${\mathbb{F}}_2[t]$, $|t|=1$. If $p$ is odd, it is $\Lambda (a)\otimes {\mathbb{F}}_p[b]$, with $|a|=1$, $|b|=2$: graded commutativity gives $a^2=0$ and the periodicity class $b$ induces the degree-two isomorphisms. The Künneth theorem gives two copies of these generators for $C_p\times C_p$; the dimension in degree $d$ is $d+1$ for every $p$.

First use the original null coordinates $u=t+m/\lambda$, $v=t-m/\lambda$. They obey $u-v=2m/\lambda$, and the radial metric tensor becomesApply the given substitutions $u=\tan (U/2)$ and $v=-\cot (V/2)$. The identitiesgive the analytic extension of a spacetimeThe poles of the individual tangent and cotangent substitutions are now ordinary coordinate lines wherever $\cos ((U-V)/2)\ne 0$. Set $T=(U+V)/2$ and $X=(U-V)/2$. Allowing these coordinate functions to continue across every such line givesThis is the universal cover of two-dimensional anti-de Sitter spacetime of radius $m$, times a two-sphere of radius $m$: the simply connected Bertotti-Robinson spacetime. In defining this maximal analytic extension, $T$ ranges over the whole real line without periodic identification. A periodic identification would instead create closed timelike curves and would be an additional global choice.

For clarity, the original patch is embedded by the branches $-\pi <U<\pi$ and $0<V<2\pi$. The inequality $u>v$ is equivalent on these branches to $|U-V|<\pi$, so the original patch occupiesThe first spacetime extension in part (a) occupies the larger band $0<T-X<2\pi$, with $|X|<\pi /2$. Its finite-$v$ Killing horizon is the interior line $T+X=\pi$. The edges $T-X=0,2\pi$ of that band are further regular Poincaré horizons; continuing through them produces the full conformal strip of two-dimensional anti-de Sitter spacetime.

The original coordinate $\lambda$ has a smooth continuation throughout the strip:Consequently the complete set of zero-level Killing horizons isThese are two alternating families of parallel null lines; different families meet only at the excluded conformal boundaries, never inside the spacetime. The continuation of the original Killing vector field isThe field never vanishes in the interior, including on the Killing horizons.

To obtain the Penrose diagram, suppress the two-sphere and multiply by the conformal factor ${\cos }^{2}X/m^2$. The resulting radial metric tensor is $-d{T}^2+d{X}^2$, so the diagram is an infinitely tall strip of width $\pi$, with null lines at 45 degrees and timelike edges $X=\pm \pi /2$. Generically,At the special edge points where the two families of Killing horizons end, the displayed formula is a $0/0$ limit and $\lambda$ has no unique limiting value. Thus the infinity labels on the diagram refer to the open edge segments, not those vertices. Each interior point of the diagram represents a two-sphere of fixed radius $m$; neither edge is a centre where that two-sphere shrinks.

Finally, these timelike edges are conformal boundaries, not omitted regular points of a further spacetime extension. One way to establish maximality is geodesic completeness. The two-dimensional anti-de Sitter spacetime factor embeds locally as $Z\cdot Z=-m^2$ in a flat space with metric signature $(-,-,+)$. Along an affinely parametrized geodesic with constant $ϵ=\dot{Z}\cdot \dot{Z}$, the ambient equation isIts oscillatory, linear or hyperbolic solutions for negative, zero or positive $ϵ$ exist for all real affine parameters and lift to the universal cover. The two-sphere is also geodesically complete, and geodesics of the product split into geodesics of its factors. Hence $\tilde{g}$ is geodesically complete. A proper regular spacetime extension would furnish a geodesic reaching its boundary in finite affine parameter, contradicting that completeness. This proves that the displayed simply connected analytic spacetime is maximal.

The uniformization theorem says that every simply connected Riemann surface is biholomorphic to the Riemann sphere, the complex plane, or the unit disc. Its consequence for a compact Riemann surface of genus at least two is that the universal cover is the unit disc, equivalently the hyperbolic plane.

To explain the restriction, a spherical universal cover would make the fundamental group finite. Since a finite covering map multiplies Euler characteristic, $2=m(2-2g)$ is impossible for $g\ge 2$. If the universal cover were the complex plane, its deck transformations would be biholomorphisms $z\mapsto az+b$. A nonidentity such map with $a\ne 1$ has a fixed point, impossible for a deck transformation. Thus the deck transformation group would consist of translations. A discrete subgroup of translations with compact quotient space is a rank-two Euclidean lattice, whose quotient space is a torus, again incompatible with $g\ge 2$.

On the unit disc, the Poincare metric on a Riemann surfacehas Gaussian curvature $-1$ and is invariant under every biholomorphism of the unit disc. It therefore descends through the deck transformation group to a hyperbolic metric on $X$. Its local expression is a positive multiple of $|dz{|}^2$ in every holomorphic coordinate chart, so it is compatible with the given complex structure. In fact, this hyperbolic metric is unique within that conformal class.

The universal cover of $S^1\times S^2$ isBecause $S^3$ is simply connected, any continuous map $f:S^3\to S^1\times S^2$ lifts to $\tilde{f}:S^3\to \mathbb{R}\times S^2$ after choosing a lift of one base point. The lifting criterion for a covering space applies because $f_{\ast }{\pi }_1(S^3)=0$. Equivalently, path lifting defines the lift and simple connectedness makes its value independent of the chosen path.

The cover deformation retracts onto $\{0\}\times S^2$, so $H_3(\mathbb{R}\times S^2;\mathbb{Z})=0$. Since $f=p\circ \tilde{f}$, functoriality of homology givesTherefore $\boxed{f_{\ast }=0\text{ and }\mathrm{deg}f=0}$, with the last equality using the oriented three-dimensional fundamental classes.

Put $\Gamma ={\pi }_1(X,x)$ and $R=\mathbb{Z}[\Gamma ]$. Give $\mathbb{Z}$ its trivial left $R$-module structure, through the augmentation $\epsilon :R\to \mathbb{Z}$, $\sum n_{g}g\mapsto \sum n_g$. The universal cover $\tilde{X}$ has the lifted cell structure, with $\Gamma$ acting by deck transformations.

For each cell of $X$, choose one lifted cell and orient every translate compatibly with its projection. The translates form one copy of the regular $R$-module. Thusis a free left $R$-module, and its boundary is $R$-linear. The augmentation $C_0(\tilde{X})\to \mathbb{Z}$ sends every vertex to one. The contractibility of $\tilde{X}$ says that its reduced homology vanishes, so the augmented complexis exact. It is therefore a free resolution of the trivial module. By the definition of the Ext functor,

An $R$-linear cochain with values in the trivial module has the same value on every translate of a lifted cell. It is therefore exactly an ordinary integral cellular cochain on $X$. The correspondence does not depend on which lift was initially chosen. It also respects the coboundary: applying a constant-on-orbits cochain to the lifted cellular boundary sums precisely the incidence coefficients of the boundary in the base. Henceas cochain complexes, not just as graded groups. Cellular cohomology computes the cohomology of a cell complex, givingAt $i=0$ both sides are $\mathbb{Z}$, by connectedness and ${\mathrm{Hom}}_R(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$. This is the cellular free resolution from a contractible universal cover description of group cohomology.