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A three-manifold is a topological manifold locally modeled on ${\mathbb{R}}^3$, or on a half-space at its boundary.

The category of topological manifolds has topological manifolds without boundary as its objects of a category and all continuous maps between them as its categorical morphisms. Allowing every dimension of a manifold lets maps connect spaces of different dimensions. Identity functions are continuous, and function composition preserves continuity, so the category laws are inherited from functions. Its isomorphisms are exactly homeomorphisms.

In coordinate charts $(U,\phi )$ on $M$ and $(V,\psi )$ on $N$, a continuous map $f:M\to N$ is represented locally by $\psi \circ f\circ {\phi }^{-1}$ on $\phi (U\cap f^{-1}(V))$. Changing coordinate charts changes this expression through topological transition maps, while $f$ itself stays fixed. This viewpoint separates the underlying continuous map from the coordinates used to describe it; restricting to smooth manifolds and smooth maps gives the corresponding smooth category.

A topological manifold has dimension $n$ when its manifold charts have coordinate domains open in ${\mathbb{R}}^n$. For a smooth manifold, each tangent space has dimension $n$, and a tangent bundle has vector bundle rank $n$.

A foliation decomposes a topological manifold into connected leaves which locally look like parallel slices in product coordinate charts.

Invariance of domain says that a continuous map that is an injective function between topological manifolds without boundary of the same dimension is an open map. It is useful in the Penrose singularity theorem: projecting a compact achronal boundary along timelike curves onto a Cauchy hypersurface gives a nonempty subset that is both open and closed, forcing that connected Cauchy hypersurface to be compact.

For a nonempty compact topological manifold $X^d$ without boundary, its suspension is a topological manifold without boundary exactly when $X$ is homeomorphic to $S^d$. The forward implication is a local homology from a link calculation at a suspension vertex: ${\tilde{H}}_j(X)$ is $\mathbb{Z}$ in degree $d$ and zero elsewhere. For $d\ge 2$, put a coordinate ball between two nested cone neighbourhoods. The inclusion of their punctures is a homotopy equivalence but factors through the simply connected punctured ball, forcing ${\pi }_1(X)=0$. Thus $X$ is a homotopy sphere, and the topological generalized Poincare theorem gives the result. The dimensions zero and one follow from the finite-set and circle classifications. Conversely $S(S^d)\cong S^{d+1}$. If manifolds with boundary are allowed, the interval and its disk suspension show why this formulation needs the boundary restriction.

A topological space is paracompact when every open cover has a locally finite open refinement. Every metric space and every second-countable topological manifold is paracompact.

Choose a generator $[S^n]$ of ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree of a continuous map is the integer determined byFor $n\ge 1$ this is ordinary top-dimensional homology; using reduced homology also gives a convention for $n=0$. Domain and target use the same orientation, so reversing the chosen generator in both leaves the integer unchanged. Homotopy invariance of homology makes the mapping degree invariant under homotopy.

We interpret the manifold criterion for a suspension in the category of nonempty topological manifolds without boundary, and “sphere” as homeomorphic to the standard sphere. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit homeomorphism from its suspension to $S^{d+1}$ isIt is a continuous map that is a bijection: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The compact-to-Hausdorff continuous bijection theorem makes it a homeomorphism.

Conversely suppose $SX$ is a topological manifold. Its open part $X\times (0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is contractible, while $U\setminus \{v\}$ admits a deformation retraction onto $X$. The long exact sequence in relative homology and excision give the local homology from a link calculationA point of a $(d+1)$-dimensional topological manifold without boundary has local homology $\mathbb{Z}$ in degree $d+1$ and zero in all other degrees. ConsequentlyThus $X$ is an integral homology sphere. This conclusion alone does not identify its homeomorphism type.

If $d\ge 2$, it is also simply connected. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By compactness of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times \{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor asThe composite induces an isomorphism of fundamental groups, since both cone punctures are products of $X$ with intervals and their inclusion is a homotopy equivalence. The middle punctured ball has the homotopy type of $S^d$ and is simply connected for $d\ge 2$. Hence the composite is also zero on the fundamental group, forcing ${\pi }_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a sphere.

The Hurewicz theorem, applied successively to the vanishing lower homology groups, now gives ${\pi }_j(X)=0$ for $j<d$ and ${\pi }_d(X)\cong H_d(X)\cong \mathbb{Z}$. A representative of its generator is a continuous map $S^d\to X$ inducing an isomorphism in every integral homology degree. Topological manifolds have the homotopy type of CW complexes, so the homological Whitehead theorem makes it a homotopy equivalence. We have proved that $X$ is a homotopy sphere.

At this point the topological classification theorem is essential: the topological generalized Poincare theorem says that a closed homotopy sphere is homeomorphic to the standard sphere. Applying it completes the converse for $d\ge 2$. For $d=1$, the local calculation makes $X$ a connected compact one-dimensional manifold without boundary, hence a circle. For $d=0$, a compact zero-dimensional manifold is a finite set, and ${\tilde{H}}_0(X)\cong \mathbb{Z}$ forces exactly two points. Thus it is $S^0$. Altogether,The last classification step is not a consequence of homology alone, and we make no assertion of diffeomorphism with a prescribed smooth structure. If the word manifold allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto ((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\le 1$, although the interval is not a sphere.

To construct sphere maps of arbitrary integer degree, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta \mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\mathrm{deg}{h}_m=m$.

Two cone neighbourhoods cover a suspension; their intersection admits a deformation retraction onto the original space. The reduced Mayer–Vietoris theorem gives a natural suspension isomorphismNaturality means that the suspension of a continuous map induces the same integer multiplier. This proves degree under suspension. Using the standard identification $S(S^k)\cong S^{k+1}$, setThus every integer occurs in every positive sphere dimension, with negative degrees handled just as well as positive ones.

If “every $n$” includes zero, that final printed assertion is false. The generator of ${\tilde{H}}_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving preciselyThese degrees of maps of the zero-sphere are the explicit exception to the arbitrary-degree construction.

An $n$-dimensional smooth manifold is a Hausdorff space with a countable topological base, equipped with a smooth atlas of homeomorphisms from open subsets onto open subsets of ${\mathbb{R}}^n$, whose overlap maps are smooth diffeomorphisms. The smooth structure is the maximal smooth atlas compatible with these manifold charts. Here manifolds have no boundary unless specified otherwise.

For a space with all the requested topological properties but no such atlas, take the topological tripod: three closed intervals joined at one endpoint. It is a compact connected subspace of the plane with its induced metric. A countable base of planar rational balls restricts to a countable base, the metric makes it Hausdorff, and compactness gives a finite subcover of every open cover, hence a locally finite refinement and paracompactness.

At an interior point of any arm, arbitrarily small neighborhoods are intervals. A coordinate ball in dimension at least two would remain connected after deleting its center; an interval does not. Dimension zero would make the space discrete. Thus any possible connected manifold structure would have dimension one. But a sufficiently small neighborhood of the junction, minus the junction, has three components, whereas an interval chart has two. This contradiction rules out even a topological manifold structure, and hence any smooth one.

The product smooth structure uses manifold charts $(\varphi ,\psi ):U\times V\to \varphi (U)\times \psi (V)\subset {\mathbb{R}}^{m+n}$. Transition maps act separately in the two coordinate blocks and are smooth with smooth inverses. Products of countable bases give a countable base, and the product remains Hausdorff. Therefore $\boxed{\dim (M\times N)=\dim M+\dim N}$ with this natural smooth structure.

Define the tangent space by point derivations: a tangent vector at $x$ is an $\mathbb{R}$-linear map $D$ on germs of smooth functions at $x$, satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. Addition and scalar multiplication preserve this rule, so these derivations form a vector space. In coordinates $u^1,\dots ,u^n$, the local identityfollows by integrating the derivative of $h$ along the coordinate line segment. Derivations annihilate constants, so it gives $Dh={\sum }_{i}D(u^i){\partial }_{i}h(u(x))$. The coordinate derivations ${\partial }_i{|}_x$ are independent since they evaluate the coordinate functions as ${\delta }_i^j$. ThusFor a smooth map, define the differential of a smooth map intrinsically by $(f_{\ast }D)(h)=D(h\circ f)$. It is again linear and satisfies the derivation rule at $f(x)$, so it maps $T_{x}M$ into $T_{f(x)}N$. Its coordinate matrix is the Jacobian of the coordinate expression of $f$, independently of the charts by the intrinsic definition and chain rule.

Finally, a zero differential forces each target coordinate function to have all derivatives zero on a small connected source coordinate ball mapping into one target chart. Integration on straight segments makes those functions constant there. Hence $f$ is locally constant. Each nonempty fiber is both open and closed, so connectedness gives the zero-differential constancy theorem, $\boxed{f\text{ is constant}}$.

The relevant Penrose singularity theorem states: a connected time-oriented four-dimensional globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a nonempty compact orientable boundaryless trapped surface, and the null convergence condition is future null-geodesically incomplete. Under the Einstein field equations, the null energy condition supplies the curvature hypothesis. A trapped surface here is spacelike and has both future normal null expansions strictly negative. The conclusion is an incomplete null geodesic, not necessarily a divergent curvature invariant at an identifiable point.

Suppose, for contradiction, that every future null geodesic is complete. Write $S$ for the trapped surface andThe equality follows because the causal future of a compact set is closed in a globally hyperbolic spacetime. This future horismos is a closed achronal boundary. It is nonempty: the restriction of a Cauchy time function to compact $S$ has a minimum, and a point at that minimum cannot be chronologically preceded by another point of $S$.

Normalize the two future null normal directions along $S$ against a smooth future timelike field, fixing the affine scale continuously. Their initial null expansions are continuous and strictly negative. Compactness of the normalized normal bundle gives a uniform $c>0$ with ${\theta }_{\pm }\le -c$ everywhere on $S$. The preceding null focusing theorem forces a conjugate point to a spacelike surface along each normal generator within affine length $L=2/c$.

By the supplied boundary-generator result, every point of $E^{+}(S)$ is reached by an orthogonal future null geodesic which has no earlier conjugate point. No such boundary generator can remain on the boundary beyond its first focal point. Every point of $E^{+}(S)$ therefore lies inThe parameter set is compact. Future null completeness makes its geodesic flow defined throughout this common finite interval, and smooth dependence on initial data makes its image $K$ compact. Since $E^{+}(S)$ is closed and contained in $K$, it is compact. This is the compactness of the future horismos of a trapped surface step; a merely pointwise finite bound would not suffice without compactness and uniform normalization.

Choose a smooth complete timelike vector field, obtained if necessary by positive rescaling against a complete auxiliary Riemannian metric. Its inextendible integral curves of a vector field meet a chosen smooth Cauchy hypersurface $C$ exactly once. Projection along those curves defines a continuous map ${\pi }_C:E^{+}(S)\to C$. The achronal set property makes this map an injective function: two boundary points on the same timelike integral curve of a vector field would be timelike related.

An achronal boundary is a topological hypersurface without boundary, even at nonsmooth generator junctions; the submanifold property allowed in the question supplies this fact. Thus domain and codomain are both three-dimensional topological manifolds without boundary. Invariance of domain makes ${\pi }_C(E^{+}(S))$ open in $C$. Compactness makes it closed in the Hausdorff space $C$, and it is nonempty. Connectedness of the spacetime gives connectedness of $C$, so this image is all of $C$. That would make $C$ compact, contradicting the noncompact Cauchy hypersurface hypothesis. The completeness assumption is false, proving the theorem.

The noncompactness hypothesis and strict trapping are essential to this version. Global hyperbolicity cannot simply be omitted in the compactness and projection steps, and nonpositive initial expansion with zeros does not supply the uniform focusing bound used here.

A complex manifold of complex dimension $n$ is a Hausdorff, second-countable topological manifold with charts to open subsets of ${\mathbb{C}}^n$ whose transition maps are biholomorphic. A rank-$r$ holomorphic vector bundle is a complex vector bundle with local trivializations $E{|}_U\cong U\times {\mathbb{C}}^r$ whose transition matrices are holomorphic maps to ${\mathrm{GL}}_r(\mathbb{C})$.

A smooth atlas is a collection of manifold charts covering a topological manifold whose coordinate changes are smooth transition maps. It determines a smooth manifold structure by adjoining every compatible chart.

A topological atlas is a family $\mathcal{A}=\{(U_i,{\phi }_i)\}$ of coordinate charts whose domains form an open cover of a topological manifold. Each point must belong to at least one chart domain; no individual domain needs to cover the entire topological manifold. On overlaps, topological transition maps describe the same points in two coordinate systems.

Every closed topological manifold with the homotopy type of $S^d$ is homeomorphic to $S^d$. This is a classification theorem about homeomorphisms, and does not assert a diffeomorphism between prescribed smooth structures. It is the precise final step needed after proving that a cone link is a simply connected integral homology sphere in the manifold criterion for a suspension.

A topological $n$-manifold without boundary is a topological space $M$ satisfying three conditions: it is Hausdorff, it is second countable, and every point $p\in M$ lies in an open set $U\subseteq M$ with a homeomorphism $\phi :U\to V$ onto an open set $V\subseteq {\mathbb{R}}^n$. Here $n\ge 0$ is a fixed dimension of a manifold, and ${\mathbb{R}}^n$ has the usual topology of Euclidean space. The pair $(U,\phi )$ is a coordinate chart. Unless explicitly qualified, “topological manifold” here means without boundary.^[ref]

The Hausdorff condition requires disjoint open neighbourhoods for distinct points. Second countability requires a countable basis of a topology. Neither follows merely from the local homeomorphisms: non-Hausdorff manifolds fail the first condition, while an uncountable disjoint union of copies of the real line fails the second. Indeed, a basis of a topology must supply a different nonempty member inside each copy, so it cannot be countable.

The circle and the real line are both one-dimensional topological manifolds, with the same local model. Their global topologies differ: the circle is compact, while the real line is not. By contrast, the junction of a topological tripod already fails the local test. Thus local coordinate charts specify what kind of small pieces are allowed; global topological invariants distinguish how those pieces fit together.^[ref]

The usefulness of a topological manifold is that local questions about continuity can be transported into Euclidean space, while compactness, connectedness, and the fundamental group still record global information. For calculus, further structure is needed: a smooth atlas requires smooth transition maps, letting the chain rule reconcile derivatives computed in different coordinate charts. A topological atlas by itself only guarantees homeomorphisms between coordinates.^[ref]