Topology

Topology studies topological spaces through open sets and continuous maps. Its central equivalence is a homeomorphism: a bijection that preserves continuity in both directions. This lets us distinguish spaces by connectedness or compactness even when numerical metric values are irrelevant. The open interval $(0,1)$ and the real line have the same topology in this sense, although only the former is a bounded set with its usual metric.

The upper panels show an explicit homeomorphism and the inverse image test for a continuous map. The lower panels exhibit an obstruction to a homeomorphism: removing any point from a circle leaves a connected space, whereas removing an interior point from a closed interval gives two connected components.

Homotopy is a different notion: it varies a continuous map continuously with a parameter and may collapse points. For example, $H(x,t)=(1-t)x$ contracts $[0,1]$ to zero. Thus this closed interval and a singleton have the same homotopy type, but there is no homeomorphism between them because there is no bijection. Homotopy equivalence need not preserve compactness. ^[ref]

The abstract viewpoint already appears explicitly in Tietze's 1923 paper: he describes topology as the study of properties invariant under invertible continuous maps with continuous inverses, and discusses axioms for neighbourhoods, citing Hausdorff's 1914 book, page 213. ^[ref]

For a family of functions $f_i:X\to Y_i$ into topological spaces, the initial topology on $X$ is the coarsest topology making them all continuous. Its subbasis consists of $f_i^{-1}(U)$ for open subsets $U\subseteq Y_i$. A map into $X$ is continuous exactly when all its composites with the $f_i$ are continuous.

On the real plane, take the initial topology of all separately continuous real-valued functions. Its coordinate lines have their usual induced topology, and the nonzero rational grid is dense, but its diagonal is uncountable and discrete. Thus the space is separable and nonmetrizable. Rectangular bump functions show that no sequence from the nonzero rational grid converges to the origin, so the space is not first countable either.

Fix a point $a$ of an uncountable set $X$ and declare $F$ closed when $a\in F$ or $F$ is countable. Finite unions and arbitrary intersections satisfy the closed-set axioms. Every convergent sequence is eventually constant: the countable set of its values different from the limit is closed, so its complement is a neighborhood of the limit. A real-valued map constant at $f(a)$ outside a countable set is continuous, because each closed inverse image either contains $a$ or lies in that countable exceptional set.

A space is zero-dimensional when its topology has a base of clopen sets. In the Hausdorff setting these sets separate distinct points, so the space is totally disconnected. Conversely a compact Hausdorff totally disconnected space has a clopen base, by the Šura-Bura component lemma and finite subcovers of the complement of a neighborhood.

The characteristic functions of a clopen topological basis give a topological embedding of a zero-dimensional $T_0$ topological space into a product of discrete two-point topological spaces. The separation axiom makes those functions distinguish points, and their coordinate preimages recover the original topological basis.

In a normal Hausdorff space, strong zero-dimensionality means that any disjoint closed sets can be separated by a clopen set, equivalently by disjoint clopen neighborhoods. For general Tychonoff spaces, the usual definition asks this for completely separated sets. In the normal setting Urysohn's lemma makes the formulations agree. The condition is equivalent to zero-dimensionality of the Stone-Čech compactification.

For normal Hausdorff $X$, if $\beta X$ has a clopen base, extend a Urysohn's lemma separator of two closed subsets of $X$. Their closures in $\beta X$ are separated; compactness gives a clopen separator, whose trace works in $X$. Conversely separate the traces of the low and high level sets of a continuous $f:\beta X\to [0,1]$ by clopen $U\subseteq X$. The characteristic function of $U$ extends into $\{0,1\}$ on $\beta X$. Density ensures its one level contains the zero level of $f$ and misses the one level. Applying this with $f$ separating a point from a closed set gives a clopen base for $\beta X$.

A Hausdorff compactification is a compact Hausdorff space $K$ with a topological embedding $j:X\to K$ whose image is dense. It retains the original topology on $X$ while adjoining boundary points. For a locally compact noncompact Hausdorff space, its one-point compactification is an example. For every Tychonoff space, the Stone-Čech compactification extends all continuous maps into compact Hausdorff targets.

If a locally compact Hausdorff space is densely embedded in a Hausdorff space, its image is open. Around a point choose an ambient open set whose trace lies in a compact neighborhood $K$. Density puts that ambient open set inside the ambient closure of its trace, which lies in $K$ because compact subsets of a Hausdorff space are closed. Thus the ambient open set lies in the image.

For a Hausdorff compactification $h:\mathbb{R}\to Y$, let $C^{\pm }$ be the intersections of the closures of the positive and negative tails, respectively. Each is nonempty and connected by the closed-filter-base theorem. Their union equals the remainder: an image point has a bounded-interval neighbourhood excluding sufficiently long tails, whereas every neighbourhood of a remainder point meets tails by density and compactness of bounded intervals. If the remainder is finite, each $C^{\pm }$ must be a singleton, so there can be only one or two remainder points. In particular there is no three-point compactification of the real line.

A Tychonoff space is Čech-complete if it is a G-delta set in a Hausdorff compactification. The G-delta compactification invariance makes this independent of the chosen compactification, and equivalent to being a G-delta set in its Stone-Čech compactification.

Write the dense space $X$ as ${\bigcap }_n{G}_n$ in compact Hausdorff $K$. Every $G_n$ is dense and open. A dense open subset $U_n$ of $X$ has the form $X\cap O_n$ with $O_n$ dense and open in $K$. The compact Baire category theorem makes ${\bigcap }_n(G_n\cap O_n)={\bigcap }_n{U}_n$ dense in $K$, hence in $X$. Therefore every Čech-complete space is a Baire space. The rationals are not a Baire space, as deleting their points one at a time gives dense open sets with empty intersection.

For the comparison map $\pi :\beta X\to K$, the fibers of compactification comparison maps identity ${\pi }^{-1}(X)=X$ pulls a G-delta set presentation of $X$ back from $K$. Conversely if $\beta X\setminus X={\bigcup }_n{F}_n$ with $F_n$ closed compact, then $K\setminus X={\bigcup }_n\pi (F_n)$. Each image is closed and misses $X$. Thus $X$ is a G-delta set in one Hausdorff compactification exactly when it is in every one.

If $\pi :\beta X\to K$ is the canonical map from the Stone-Čech compactification onto a Hausdorff compactification, its fibers over the original points are singletons. If $\pi (z)=x\in X$ and $z\ne x$, choose an open neighborhood $U$ of $x$ in $\beta X$ whose closure excludes $z$. Choose open $V\subseteq K$ with $x\in V$ and $V\cap X\subseteq U$. The open neighborhood ${\pi }^{-1}V$ of $z$ has dense trace $V\cap X$, placing $z$ in $\overline{U}$, a contradiction. This controls when images of closed remainder sets stay in the remainder.

For two topologies on the same set $X$, $\sigma$ is finer than $\tau$ when every $\tau$-open set is $\sigma$-open. Equivalently, the identity function $(X,\sigma )\to (X,\tau )$ is continuous, by the definition using preimages of open sets. The inclusion may be an equality; strictly finer means proper inclusion.

The topological disjoint union of topological spaces $X_i$ has the labelled disjoint union as its underlying set. A subset is open exactly when its intersection with each labelled copy $X_i$ is open in that copy. Each copy is a clopen set, and a function from the topological disjoint union is continuous exactly when its restriction to every copy is continuous. This follows immediately by testing preimages of open sets on each copy. A topological space is homeomorphic to the topological disjoint union of its connected components exactly when all those connected components are open.

A topological invariant is a property or mathematical object preserved by homeomorphisms of topological spaces. Connectedness and compactness are examples. The fundamental group is an invariant up to group isomorphism; it is also preserved up to isomorphism by homotopy equivalence, a stronger invariance property which does not hold for every topological invariant.

An end of a noncompact topological space describes a coherent way to escape every compact subset. For a cofinal compact exhaustion, when one exists, $K_1\subseteq K_2\subseteq \cdots$, it is represented by nested connected components of the complements. A connected noncompact manifold is disconnected at infinity when it has more than one end.

Orientability asks whether local choices of handedness can agree everywhere on a manifold. On a smooth manifold, an orientation of a smooth manifold chooses an orientation of a vector space for each tangent space, continuously with position. An ordered basis describes the same handedness as another one when the change-of-basis determinant is positive. An oriented atlas implements this condition through positive Jacobian determinants of the smooth transition maps.

A paper strip makes the obstruction visible. Join its short ends without twisting: the resulting untwisted band admits a consistent choice of unit normal. Add a half-turn before joining: on the Möbius band, following the core circle of a Möbius band reverses that choice when you first return to the same point. A local decision about handedness has encountered a global contradiction. The normal reversal around the Möbius core makes this precise for an explicit model.

Both rectangles have coordinates $(u,v)\in [0,1]\times [-1,1]$. The untwisted band identifies $(0,v)$ with $(1,v)$; the Möbius band identifies $(0,v)$ with $(1,-v)$. Across a seam the corresponding local coordinate changes preserve or reverse $v$, giving determinants $+1$ and $-1$. This sign explains the handedness change; the drawing also tracks the two boundary circles of the untwisted band and the single boundary circle of the Möbius band.

For a smooth topological surface embedded in ${\mathbb{R}}^3$, choosing an orientation of a surface is equivalent to choosing a continuous unit normal. This picture is specific to surfaces in three-dimensional space: the general definition uses compatible local orientations, without requiring an embedding. Orientability is a topological invariant, so a homeomorphism cannot turn a Möbius band into a untwisted band. The sphere and torus are orientable surfaces; the Klein bottle is not. ^[ref]

The distinction matters when a sign has geometric meaning. A signed flux surface integral needs an orientation of a surface, and Stokes theorem relates that choice to the outward-normal-first boundary orientation. Ordinary unsigned surface area does not require orientability. Thus a Möbius band still has a perfectly meaningful area even though it has no continuous global unit normal.

A connected orientable manifold of positive dimension has two possible orientations: reversing every local choice gives the other one. If a manifold has several connected components, each component can be oriented independently when it is orientable. For topological surfaces, a useful obstruction criterion is the presence of a Möbius band: a surface is nonorientable exactly when it contains such a band. ^[ref]

A non-Hausdorff manifold is a locally Euclidean topological space for which distinct points need not have disjoint neighborhoods. The real line with two origins is a basic example. Nonseparated algebraic schemes exhibit analogous gluing phenomena, but are not generally topological manifolds.

Take $\mathbb{R}\times \{0,1\}$ with its Euclidean topology and identify $(t,0)$ with $(t,1)$ for every $t\ne 0$, retaining two distinct origins. This quotient space is locally homeomorphic to the real line but is not Hausdorff. Any open neighbourhoods of its origins have preimages containing $(-\epsilon ,\epsilon )\times \{0\}$ and $(-\delta ,\delta )\times \{1\}$, so a nonzero $|t|<\min (\epsilon ,\delta )$ gives a shared image point. All equivalence classes in the original metric space are finite and closed. This shows that closed equivalence classes do not suffice for a Hausdorff quotient space.

A Stone space is a compact Hausdorff space that is totally disconnected, equivalently a compact Hausdorff space with a basis of clopen sets. The topology on a type space gives an important example.

The nonempty open sets are the upper tails of the natural numbers. Arbitrary nonempty unions select the smallest starting index, and finite intersections select the largest, so these sets form a topology. A map into this space is continuous exactly when the inverse image of every upper tail is open.

For a map $f:\mathbb{N}\to \mathbb{N}$ from the cofinite topology to the tail topology on the natural numbers, continuity is equivalent to every set $\{n:f(n)\ge k\}$ being empty or cofinite. If the image is unbounded, every such set is nonempty and cofinite, so $f(n)\to \infty$. If it is bounded, its maximum $M$ exists, and the inverse image of the tail at $M$ forces $f(n)=M$ eventually, with $f(n)\le M$ always. Conversely either condition makes every tail preimage empty or cofinite.

The Fell topology on the closed subsets of a locally compact space has subbasic open sets requiring intersection with an open set or avoidance of a compact set. On a locally compact separable metric space, its sequential convergence says that every convergent subsequence of points from the varying sets has its limit in the limiting set, and every point of the limiting set is approximated by points from the varying sets.

These sets form a topology: a nonempty union of upper rays is another upper ray, or all of $\mathbb{R}$ when their endpoints have no lower bound; a finite intersection of rays has endpoint the maximum of their endpoints. The whole line must be included. The continuous map $n\mapsto n$ from $\mathbb{N}$ with the cofinite topology to this space is nonconstant, because each upper ray has cofinite inverse image.

For the topology consisting of the empty set, the whole real line, and upper open rays, a nonempty subset is compact exactly when it has a smallest element. An open set containing the smallest element contains the entire subset. Conversely, if there is no smallest element, the rays $(a,\infty )$ indexed by points $a$ of the subset cover it, but any finite selection misses its smallest chosen endpoint.

A topological space is hyperconnected if any two nonempty open sets intersect. Every continuous map from a nonempty hyperconnected space to a Hausdorff space is constant: two distinct image points would have disjoint open neighbourhoods whose nonempty inverse images contradict hyperconnectedness. Every infinite set with the cofinite topology is hyperconnected.

A closed cover of a topological space is a family of closed sets whose union is the whole space. A finite closed cover can be used in separation and antipodal arguments: in a metric space, vanishing distance to a closed member means actual membership. Unlike an open cover, a closed cover does not automatically have a partition of unity subordinate to its members.

Given a partition of a set, declare a subset open precisely when it is a union of partition blocks. Arbitrary unions and finite intersections of such unions are again unions of blocks, giving a topology. The quotient map to the block set with its discrete topology pulls that topology back to the original set. For the partition of $\mathbb{Z}$ into its two parity classes, the only open sets are the empty set, those two classes, and all of $\mathbb{Z}$.

On an infinite set, the family consisting of all finite subsets and the whole set need not be closed under arbitrary unions. On $\mathbb{Z}$, the union of the singleton subsets of $2\mathbb{Z}$ is infinite and proper, so this family is not a topology. It must not be confused with the cofinite topology, whose proper open sets have finite complements.

The lower limit topology on $\mathbb{R}$ has basis of a topology $\{[a,b):a<b\}$. Every such nonempty interval contains a rational, so this topological space is separable. Its topology is finer than the usual Euclidean topology; the one-sided basic neighbourhoods allow its square to have an uncountable discrete antidiagonal.

The Sorgenfrey plane is the product topology on two copies of the Sorgenfrey line. It is separable because ${\mathbb{Q}}^2$ meets every nonempty basic rectangle $[a,b)\times [c,d)$. Its antidiagonal $\{(x,-x):x\in \mathbb{R}\}$ is an uncountable discrete subspace: $[x,x+ϵ)\times [-x,-x+ϵ)$ meets it only at $(x,-x)$. In a discrete space every dense subset is the whole space, so this subspace is not separable. Thus separability need not pass to subspaces of nonmetrizable spaces.

// Target: analysis.bigb

The $g$th symmetric product of a space $X$ is ${\mathrm{Sym}}^g(X)=X^g/S_g$, where the symmetric group permutes the factors. Its points are unordered $g$-tuples, with repetitions allowed. If $X$ is a Riemann surface, elementary symmetric functions in local complex coordinates make its symmetric product a smooth complex manifold, including along the locus of repeated points. A collection of disjoint circles in $X$ gives a product torus in this symmetric product.

A topological space is extremally disconnected when the closure of every open set is open. This concerns open sets, not arbitrary subsets. In particular, a nonisolated singleton in a Hausdorff space is closed but not open, so its closure gives a counterexample to that stronger condition.

The open sets are the empty set and subsets whose complement is countable. On an uncountable set, any two nonempty open sets intersect, so the space is not Hausdorff. On a countable underlying set this topology is discrete. The compact subsets of a cocountable space are finite, regardless of the cardinality of the underlying set.

Only finite subsets are compact in the cocountable topology. An infinite subset contains distinct $c_1,c_2,\dots$. The open sets obtained by removing the tails $\{c_j:j\ge n\}$ form an increasing open cover, but every finite selection misses a tail point. Finite sets are compact in every topology.

A family $(A_i)$ of subsets of a topological space is locally finite if every point has a neighbourhood meeting only finitely many of its members. On a compact space such a family has only finitely many nonempty members. For singleton subsets of a Riemann surface, this is the condition permitting prescribed poles of a meromorphic function without accumulation inside the surface.

The indiscrete topology on a set $X$ has only the empty set and the whole set as open sets. When $X$ has more than one point it is not a Hausdorff space. For example the quotient topology on $\mathbb{R}/\mathbb{Q}$ is indiscrete: a nonempty open saturated subset contains an interval and all its rational translates, which cover $\mathbb{R}$.

A triangulation of a surface decomposes it into triangles meeting along full edges or vertices. On a closed surface, each edge is incident with two triangle faces, so $3F=2E$. Euler characteristic connects this local counting relation with the surface's topology.

If a topological space is the union of finitely many closed subspaces and continuous maps on those subspaces agree on every overlap, they combine into a continuous map on the whole space. For a closed set in the target, its preimage is a finite union of closed sets, proving continuity.

The cofinite topology on a set $X$ declares the empty set and all cofinite subsets to be open sets: a nonempty open set may exclude only a finite set of points. There is no assumed metric. On an infinite set, this gives a particularly small collection of open sets while still making every singleton a closed set. On a finite set, every subset has finite complement, so the cofinite topology is the discrete topology.^[ref]

For an original example, use $X={\mathbb{N}}_{>0}$ and chooseBoth are open sets, and their intersection is $U\cap V=X\setminus \{1,2,3,5,7,8\}$. Thus $4,6,9,10,\dots$ are in both. Their finite exclusions cannot separate the infinite remainder. The finite intersection of any finite collection of nonempty open sets is again a nonempty open set, since its complement is a finite union of finite sets. An arbitrary nonempty union is open because its complement is contained in the finite complement of any one nonempty member.

The cofinite topology is useful for separating hypotheses that familiar metric spaces package together. An infinite cofinite topological space is a T1 space, because its singletons are closed, but is not a Hausdorff space, because two nonempty open sets cannot be disjoint sets. It is a hyperconnected space and hence a connected space. Moreover, every subset of a cofinite space is compact, and sequence convergence in the cofinite topology can have many limits. These are compatible properties, rather than contradictions: the usual uniqueness and closedness results use the Hausdorff space hypothesis.^[ref]

Let $Z$ be the underlying set of real numbers with the cofinite topology. The diagonal $W=\{(t,t):t\in \mathbb{R}\}$ in $Z\times \mathbb{R}$ with its subspace topology is homeomorphic to the usual real line. Indeed, $t\mapsto (t,t)$ is a continuous bijection: its first coordinate is continuous because usual finite sets are closed, and its second coordinate is the identity. The inverse is the second coordinate projection restricted to $W$. Consequently $W$ is not compact, whereas $Z$ is compact by the fact that every subset of a cofinite space is compact. Thus $W$ is not homeomorphic to $Z$.

For a set $X$ with its cofinite topology, a subset $S$ is a connected subset exactly when $S$ is empty, a singleton, or infinite. Indeed, its subspace topology is the cofinite topology on $S$: the traces of open sets are precisely $S\setminus F$ for finite $F\subseteq S$, together with the empty set. For infinite $S$, two nonempty open sets intersect, because their complements have finite union. Hence $S$ is a hyperconnected space and therefore a connected space. A finite $S$ has the discrete topology, which admits a separation of a topological space as soon as $S$ has at least two points.

For an infinite set $X$ with its cofinite topology, every infinite subset is connected, but $X\times X$ has an infinite disconnected topological subspace. Choose $a\ne b$ and use $D=\{a,b\}\times X$. The open sets $(X\setminus \{b\})\times X$ and $(X\setminus \{a\})\times X$ have disjoint nonempty traces $\{a\}\times X$ and $\{b\}\times X$ on $D$, covering it. Thus $D$ is a product of a discrete two-point subspace with an infinite cofinite factor. The ambient open rectangles may intersect; only their traces must be disjoint for a separation of $D$.

By convergence of a sequence in a topological space, $x_n\to a$ means that every open neighbourhood of $a$ contains all sufficiently late terms. In the cofinite topology, this is equivalent to the condition that, for every $b\ne a$, the value $b$ occurs only finitely often. Necessity uses the open neighbourhood $X\setminus \{b\}$; sufficiency chooses an index beyond every last occurrence of the finitely many points excluded by a given open neighbourhood.

Consequently a sequence with distinct terms converges to every point of $X$. For the original choice $x_n=n$ on $X={\mathbb{N}}_{>0}$, take the two open neighbourhoodsAll terms with $n\ge 9$ lie in $U_1$ and all terms with $n\ge 8$ lie in $U_2$. More generally, each finite excluded set has a last visit, so $x_n\to 1$ and $x_n\to 2$, and in fact $x_n\to a$ for every $a\in X$. No numerical distance is tending to zero here; the open sets specify the meaning of convergence.

Distinct terms are sufficient, not necessary. The sequence $1,1,2,2,3,3,\dots$ also converges to every point, since each value occurs only twice. By contrast, the constant sequence $x_n=1$ converges only to $1$: an open neighbourhood of any $a\ne 1$ can exclude $1$. The alternating sequence $1,2,1,2,\dots$ converges to no point, since some value different from any proposed limit occurs infinitely often.

Every subset $A$ of a cofinite topological space is a compact subset. To see the finite reduction directly, take an open cover of a nonempty $A$ by open sets of $X$ and choose one member $U$ containing a point of $A$. Since $U$ has finite complement, the set $A\setminus U$ is finite. Choose a covering member for each of those remaining points. Those members together with $U$ are a finite subcover. The empty set is already compact.

Equivalently, the subspace topology on $A$ is itself cofinite: for finite $F\subseteq X$, the trace of $X\setminus F$ is $A\setminus (F\cap A)$. The argument applies again inside $A$. A concrete open cover $\{X\setminus \{k\}:k\in X\}$ of $X={\mathbb{N}}_{>0}$ needs only the two members $X\setminus \{1\}$ and $X\setminus \{2\}$ when these labels are distinct: each fills the other's single gap.

On $X={\mathbb{N}}_{>0}$, the subset of even integers is compact but not closed, because it is infinite and proper. A compact subset of a Hausdorff space is guaranteed to be closed, whereas compactness without the Hausdorff space assumption does not suffice. Changing the topology changes compactness: the same infinite set with its discrete topology has the singleton open cover, which has no finite subcover. The finite-exception argument is special to finite complements; in the cocountable topology, compact subsets of a cocountable space must instead be finite.

If $X$ is an infinite set with its cofinite topology and $Y$ is a Hausdorff space, every continuous map $f:X\to Y$ is a constant function. Otherwise two distinct image points would have disjoint open neighbourhoods. Their inverse images would be nonempty disjoint open sets of $X$, contradicting its hyperconnected space property.

For example, $f(n)=n$ from ${\mathbb{N}}_{>0}$ with its cofinite topology into the real line with its usual Euclidean metric is not continuous: the inverse image of the open interval $(1/2,3/2)$ is $\{1\}$, which is not open in the infinite domain. A continuous real-valued readout on that domain can only report one value. Finite cofinite domains are different: their discrete topology allows arbitrary functions into any topological space.

Let $Z$ be a nonempty infinite set with the cofinite topology, let $T$ be any topological space, and let $Y$ be Hausdorff. For a continuous map $F:Z\times T\to Y$, every slice $z\mapsto F(z,t)$ is constant. Thus $F(z,t)=g(t)$, where $g(t)=F(z_0,t)$ for any fixed $z_0$. The map $g$ is continuous, since $t\mapsto (z_0,t)$ is continuous into the product topology. It is unique because the coordinate projection ${\pi }_T$ is a surjection. If $F$ is a surjection, so is $g$.

For $T=[0,1]$ and $Y$ homeomorphic to the circle, a surjective $g$ cannot be injective: otherwise the compact-to-Hausdorff continuous bijection theorem would contradict the fact that the circle and interval are not homeomorphic. Hence distinct $a,b\in [0,1]$ satisfy $F(z,a)=F(z,b)$ simultaneously for all $z\in Z$.

A continuous map from a nonempty finite product space of infinite sets with their cofinite topologies to a Hausdorff space is a constant function. Fixing all but one coordinate gives a continuous map from an infinite cofinite topology, so changing that coordinate does not change the value. Any two points in a finite Cartesian product can be joined by finitely many one-coordinate changes. This proves the assertion. In particular every real-valued continuous function on $Z\times Z$ is constant when $Z$ is the real line equipped with the cofinite topology.

A nonempty Hausdorff space is the continuous image of $[0,1]$ exactly when it is compact, connected, locally connected, and second-countable.

A Peano continuum is a compact, connected, locally connected metric space. By the Hahn-Mazurkiewicz theorem, every Peano continuum is the image of a continuous surjection from $[0,1]$.

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]