The first surprise in manifold theory is that many spaces look Euclidean only when inspected through small windows. A circle is not an interval, a sphere is not a plane, and a torus cannot be flattened into the plane without cutting it. Yet near each point these spaces admit coordinates that make local questions resemble questions in $\mathbb R^n$. A topological manifold is the minimum topological framework in which this local Euclidean behaviour is made precise before any smooth, metric, or algebraic structure is added. The definition is deliberately strict. Local Euclidean charts alone do not prevent global pathologies: a space can have local coordinates but fail to separate points well, or it can require too many coordinate patches to behave like an object of ordinary geometry. The Hausdorff and second countability hypotheses exclude these failures while preserving the familiar examples. [example: A Circle Is Locally a Line] Let $S^1\subset\mathbb R^2$ be \begin{align*} S^1=\{(x,y)\in\mathbb R^2:x^2+y^2=1\}. \end{align*} Write $N=(0,1)$, and define stereographic projection from $N$ by sending a point $(x,y)\in S^1\setminus\{N\}$ to the intersection of the line through $N$ and $(x,y)$ with the horizontal axis. A point on that line has the form $(\lambda x,1+\lambda(y-1))$. Setting the second coordinate equal to $0$ gives $1+\lambda(y-1)=0$, so $\lambda=1/(1-y)$, and therefore \begin{align*} \sigma_N(x,y)=\frac{x}{1-y}. \end{align*} We compute its inverse. For $t\in\mathbb R$, the line through $N=(0,1)$ and $(t,0)$ is parametrized by $(\lambda t,1-\lambda)$. Its intersections with $S^1$ satisfy \begin{align*} (\lambda t)^2+(1-\lambda)^2=1. \end{align*} Expanding gives \begin{align*} \lambda^2t^2+1-2\lambda+\lambda^2=1. \end{align*} Subtracting $1$ and factoring gives \begin{align*} \lambda\bigl(\lambda(t^2+1)-2\bigr)=0. \end{align*} The solution $\lambda=0$ gives the north pole, so the other intersection is obtained from $\lambda=2/(t^2+1)$. Thus \begin{align*} \sigma_N^{-1}(t)=\left(\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right). \end{align*} This point lies on $S^1$ because \begin{align*} \left(\frac{2t}{t^2+1}\right)^2+\left(\frac{t^2-1}{t^2+1}\right)^2=\frac{4t^2+t^4-2t^2+1}{(t^2+1)^2}=\frac{t^4+2t^2+1}{(t^2+1)^2}=1. \end{align*} Also, \begin{align*} 1-\frac{t^2-1}{t^2+1}=\frac{2}{t^2+1}, \end{align*} so applying $\sigma_N$ to this point gives \begin{align*} \frac{2t/(t^2+1)}{2/(t^2+1)}=t. \end{align*} The formulas for $\sigma_N$ and $\sigma_N^{-1}$ are continuous, so $\sigma_N:S^1\setminus\{N\}\to\mathbb R$ is a homeomorphism. Rotating the circle moves any chosen point to $N$, so every point of $S^1$ has a neighbourhood with one-dimensional coordinates, even though the whole circle is compact and no nonempty open subset of $\mathbb R$ is compact. [/example] This example shows the guiding tension. Local coordinates make the space look linear, while global topology records how those coordinate patches are glued together. The subject begins by separating those two roles. ## Definition ### The Manifold Axioms The main definition has to say exactly which topological spaces are eligible for coordinate-based geometry. Local Euclidean behaviour supplies the small windows, but the Hausdorff and countability assumptions keep those windows from being glued into doubled points or unmanageably large topologies. The definition below records both the local Euclidean requirement and the global topological discipline needed for geometry. [definition: Topological Manifold] Let $n$ be a fixed nonnegative integer. A topological manifold of dimension $n$ is a Hausdorff, second countable [topological space](/page/Topological%20Space) $M$ such that for every point $p\in M$ there exist an [open set](/page/Open%20Set) $U\subset M$ with $p\in U$, an open set $W\subset\mathbb R^n$, and a homeomorphism \begin{align*} \varphi:U&\to W. \end{align*} [/definition] The first condition says that distinct points can be separated by disjoint open neighbourhoods; the second says that the topology has a countable basis; and the third says that $M$ is locally homeomorphic to $\mathbb R^n$. The integer $n$ is the dimension of the manifold, but a single chart is rarely enough to describe all of $M$. ### Charts and Atlases To make the phrase "locally Euclidean" usable in computations, the coordinate neighbourhood must be open in the space and must map onto an open subset of Euclidean space. The openness condition lets local topological statements transfer across the coordinate map without changing their nature. [definition: Coordinate Chart] Let $M$ be a topological manifold of dimension $n$. A coordinate chart on $M$ is a pair $(U,\varphi)$ such that $U\subset M$ is open and there is an open set $W\subset\mathbb R^n$ for which \begin{align*} \varphi:U&\to W \end{align*} is a homeomorphism. [/definition] The coordinates of a point $p\in U$ are the [real numbers](/page/Real%20Numbers) $(x_1,\ldots,x_n)=\varphi(p)$, but at this stage the coordinates only preserve topology. They do not yet allow derivatives, lengths, angles, or integration. To use coordinates in practice, the coordinate neighbourhoods must be organized as a reusable cover of the whole space. This need leads to the atlas, the object that isolates the coordinate system as a collection rather than a single chart. Most manifolds cannot be covered by one chart, and the data of many overlapping charts becomes the language in which global questions are translated into Euclidean ones. [definition: Atlas] Let $M$ be a topological manifold of dimension $n$. An atlas on $M$ is a collection of coordinate charts $\{(U_i,\varphi_i)\}_{i\in I}$ such that \begin{align*} M=\bigcup_{i\in I}U_i. \end{align*} [/definition] An atlas is a topological object here. Later, when smooth manifolds are introduced, the transition maps between overlapping charts must satisfy differentiability conditions. For a topological manifold, only continuity and continuous invertibility are required. ## Local Models and Dimension ### Dimension Invariance The definition requires a fixed integer $n$. This is not a cosmetic choice. If different points could have different Euclidean dimensions, the space would not have a stable local type, and the usual geometric constructions would not have a single rank. [quotetheorem:9910] This theorem is the reason dimension is well-defined. It prevents a point from having both a two-dimensional coordinate neighbourhood and a three-dimensional coordinate neighbourhood. The result is topological but its proof is not elementary, so it is usually treated as a foundational background theorem. There is a related convention hidden in the word "dimension." This page uses the pure-dimensional convention: a topological manifold has one fixed local dimension $n$ across the whole space. A disjoint union $S^1\sqcup S^2$ is therefore not a manifold in this sense, although each [connected component](/page/Connected%20Component) is a manifold of its own dimension. Some authors allow such spaces and call them manifolds of varying dimension, but most geometric constructions are cleaner when the dimension is fixed from the start. ### Euclidean Pieces The most basic examples come from open subsets of Euclidean space. They are not merely examples; every manifold is assembled from pieces of this form. [example: Open Sets Are Manifolds] Let $U\subset\mathbb R^n$ be open with the [subspace topology](/page/Subspace%20Topology). We show that $U$ satisfies the three manifold axioms, using the single coordinate chart \begin{align*} \operatorname{id}_U:U\to U. \end{align*} The target $U$ is open in $\mathbb R^n$ by assumption, and $\operatorname{id}_U$ is a homeomorphism because for every open set $O\subset U$, both $\operatorname{id}_U^{-1}(O)=O$ and $(\operatorname{id}_U^{-1})^{-1}(O)=O$ are open in $U$. The space $U$ is Hausdorff. If $p,q\in U$ and $p\neq q$, set $\varepsilon=\|p-q\|/3$. The sets $U\cap B(p,\varepsilon)$ and $U\cap B(q,\varepsilon)$ are open neighbourhoods of $p$ and $q$ in $U$. If some $z$ lay in both, then \begin{align*} \|p-q\|\leq \|p-z\|+\|z-q\|<\varepsilon+\varepsilon=\frac{2}{3}\|p-q\|, \end{align*} which is impossible since $\|p-q\|>0$. The space $U$ is second countable because the sets $U\cap B(a,r)$ with $a\in\mathbb Q^n$ and $r\in\mathbb Q_{>0}$ form a countable basis. Indeed, if $O\subset U$ is open and $p\in O$, then $O=U\cap G$ for some open $G\subset\mathbb R^n$, so there is $\rho>0$ with $B(p,\rho)\subset G$. Choose $a\in\mathbb Q^n$ with $\|p-a\|<\rho/4$, and choose $r\in\mathbb Q_{>0}$ with $\|p-a\|<r<\rho/2$. Then $p\in U\cap B(a,r)$, and if $x\in B(a,r)$, then \begin{align*} \|x-p\|\leq \|x-a\|+\|a-p\|<r+\frac{\rho}{4}<\frac{3\rho}{4}<\rho, \end{align*} so $x\in B(p,\rho)\subset G$. Hence $U\cap B(a,r)\subset O$. Thus $U$ is Hausdorff, second countable, and locally homeomorphic to the open subset $U\subset\mathbb R^n$ by the identity chart, so every open subset of $\mathbb R^n$ is an $n$-dimensional topological manifold. [/example] Manifolds can also be built by simple identifications. The circle is obtained by gluing the endpoints of an interval, while a torus is obtained by gluing opposite sides of a square. The definition does not mention gluing, but local charts detect exactly when the result still looks Euclidean near every point. [example: The Torus from a Square] Let $T^2$ be the quotient of $[0,1]^2$ by the identifications $(0,t)\sim(1,t)$ and $(s,0)\sim(s,1)$, and let $q:[0,1]^2\to T^2$ be the quotient map. If $0<a<1$ and $0<b<1$, choose $\varepsilon>0$ with $a-\varepsilon>0$, $a+\varepsilon<1$, $b-\varepsilon>0$, and $b+\varepsilon<1$. On the open square \begin{align*} R=(a-\varepsilon,a+\varepsilon)\times(b-\varepsilon,b+\varepsilon), \end{align*} no two distinct points are identified, so $q|_R$ gives coordinates on $q(R)$ by \begin{align*} q(x,y)\longmapsto (x-a,y-b). \end{align*} Its inverse is $(u,v)\mapsto q(a+u,b+v)$, defined on $(-\varepsilon,\varepsilon)^2$, so interior points have ordinary planar neighbourhoods. Now take a point on a vertical identified edge, say $q(0,b)=q(1,b)$ with $0<b<1$. Choose $\varepsilon>0$ with $b-\varepsilon>0$ and $b+\varepsilon<1$, and use the saturated strip \begin{align*} E=\bigl([0,\varepsilon)\times(b-\varepsilon,b+\varepsilon)\bigr)\cup\bigl((1-\varepsilon,1]\times(b-\varepsilon,b+\varepsilon)\bigr). \end{align*} Define coordinates on $q(E)$ by \begin{align*} q(x,y)\longmapsto (x,y-b)\quad\text{if }0\leq x<\varepsilon, \end{align*} and \begin{align*} q(x,y)\longmapsto (x-1,y-b)\quad\text{if }1-\varepsilon<x\leq 1. \end{align*} The two formulas agree on identified points because $(0,y)\sim(1,y)$ and both are sent to $(0,y-b)$. The inverse sends $(u,v)$ to $q(u,b+v)$ for $0\leq u<\varepsilon$, and to $q(1+u,b+v)$ for $-\varepsilon<u\leq 0$. Thus $q(E)$ is homeomorphic to $(-\varepsilon,\varepsilon)\times(-\varepsilon,\varepsilon)$. The horizontal edges are handled in the same way, using the second coordinate instead of the first. At a corner, the four corners of the square represent one point of $T^2$. For small $\varepsilon>0$, take the union of the four corner squares \begin{align*} C=[0,\varepsilon)^2\cup(1-\varepsilon,1]\times[0,\varepsilon)\cup[0,\varepsilon)\times(1-\varepsilon,1]\cup(1-\varepsilon,1]^2. \end{align*} Send $q(x,y)$ to $(X,Y)$, where $X=x$ if $0\leq x<\varepsilon$ and $X=x-1$ if $1-\varepsilon<x\leq 1$, and where $Y=y$ if $0\leq y<\varepsilon$ and $Y=y-1$ if $1-\varepsilon<y\leq 1$. These choices respect the edge identifications, so they give a well-defined homeomorphism \begin{align*} q(C)\to(-\varepsilon,\varepsilon)^2. \end{align*} Every point of $T^2$ therefore has a neighbourhood homeomorphic to an open subset of $\mathbb R^2$, so the quotient square has the local form of a two-dimensional topological manifold. [/example] ### Branching Failures The same local test rules out many tempting spaces. A branching point is worse than a bend or a self-suggesting picture: too many arcs meet there for the neighbourhood to be an interval. The obstruction is not visual sharpness but the topology of punctured neighbourhoods. [example: A Branch Point Is Not a One-Dimensional Manifold] Let $X\subset\mathbb R^2$ be the union of the three rays \begin{align*} \{(t,0):t\geq 0\}\cup\{(-t,0):t\geq 0\}\cup\{(0,t):t\geq 0\}. \end{align*} We show that the obstruction occurs only at the origin. If $p=(a,0)$ with $a>0$, choose $0<\varepsilon<a$. Then \begin{align*} X\cap B(p,\varepsilon)=\{(x,0):a-\varepsilon<x<a+\varepsilon\}. \end{align*} The map $(x,0)\mapsto x-a$ is a homeomorphism from this neighbourhood to $(-\varepsilon,\varepsilon)$. The same argument applies on the negative horizontal ray and on the positive vertical ray, using the coordinate along that ray. Now consider the origin $0=(0,0)$. For every $\varepsilon>0$, the punctured neighbourhood \begin{align*} \bigl(X\cap B(0,\varepsilon)\bigr)\setminus\{0\} \end{align*} is the disjoint union \begin{align*} \{(t,0):0<t<\varepsilon\}\cup\{(-t,0):0<t<\varepsilon\}\cup\{(0,t):0<t<\varepsilon\}. \end{align*} Each of these three sets is connected because it is homeomorphic to $(0,\varepsilon)$, and no connected subset can meet two of them without also containing the origin, since the only common [limit point](/page/Limit%20Point) joining different rays is $(0,0)$. Hence deleting the origin from any sufficiently small neighbourhood of the origin leaves three connected components. By contrast, if $I\subset\mathbb R$ is an open interval and $s\in I$, then \begin{align*} I\setminus\{s\}=(I\cap(-\infty,s))\cup(I\cap(s,\infty)), \end{align*} and these are exactly two connected components when both sides are nonempty. A homeomorphism preserves connected components after deleting the corresponding point, so no neighbourhood of the origin in $X$ is homeomorphic to an open interval. Therefore $X$ is locally a one-dimensional manifold away from the origin, but the branch point prevents $X$ from being a one-dimensional topological manifold. [/example] This failure example captures a recurring theme. A manifold is allowed to bend, wrap, and have complicated global topology, but it is not allowed to branch locally. ## Separation and Countability ### Hausdorff Separation Local Euclidean charts alone control each small coordinate patch, but they do not guarantee that different points can be separated by neighbourhoods. The Hausdorff condition supplies that basic separation principle. [definition: Hausdorff Space] A topological space $X$ is Hausdorff if for every pair of distinct points $p,q\in X$ there exist open sets $U,V\subset X$ such that \begin{align*} p\in U,\qquad q\in V,\qquad U\cap V=\varnothing. \end{align*} [/definition] In geometry, the Hausdorff condition prevents limits from splitting. A sequence or net should not converge to two different points merely because the space was glued too loosely. Without this condition, local coordinates can exist while global topology refuses to behave like a space of points. [example: The Line with Two Origins] Let $X$ be the quotient of $\mathbb R\times\{1,2\}$ in which $(t,1)\sim(t,2)$ for every $t\neq 0$, while $(0,1)$ and $(0,2)$ remain distinct. Write $q:\mathbb R\times\{1,2\}\to X$ for the quotient map, and write $0_1=q(0,1)$ and $0_2=q(0,2)$. First, $X$ is locally Euclidean. If $a\neq 0$, choose $\varepsilon>0$ with $\varepsilon<|a|$. Then the interval $(a-\varepsilon,a+\varepsilon)$ does not contain $0$, so the two copies are already identified there. The map \begin{align*} q(t,1)\longmapsto t-a \end{align*} is a homeomorphism from $q((a-\varepsilon,a+\varepsilon)\times\{1\})$ onto $(-\varepsilon,\varepsilon)$. At the origin $0_1$, take \begin{align*} U_1=q((-\varepsilon,\varepsilon)\times\{1\}). \end{align*} Its full preimage is \begin{align*} q^{-1}(U_1)=((-\varepsilon,\varepsilon)\times\{1\})\cup((-\varepsilon,0)\cup(0,\varepsilon))\times\{2\}, \end{align*} which is open in the disjoint union $\mathbb R\times\{1,2\}$. Define \begin{align*} \phi_1(q(t,1))=t. \end{align*} For $t\neq 0$, the same point of $X$ can also be written as $q(t,2)$, and the formula still gives $t$, so $\phi_1$ is well-defined. Its inverse sends $t$ to $q(t,1)$. Thus $U_1$ is homeomorphic to $(-\varepsilon,\varepsilon)$. The same construction with the second copy gives an interval neighbourhood of $0_2$. However, $0_1$ and $0_2$ cannot be separated by disjoint open sets. Let $A$ be any open neighbourhood of $0_1$ and let $B$ be any open neighbourhood of $0_2$. Since $q^{-1}(A)$ is open and contains $(0,1)$, there is $\varepsilon>0$ such that \begin{align*} (-\varepsilon,\varepsilon)\times\{1\}\subset q^{-1}(A). \end{align*} Since $q^{-1}(B)$ is open and contains $(0,2)$, there is $\delta>0$ such that \begin{align*} (-\delta,\delta)\times\{2\}\subset q^{-1}(B). \end{align*} Choose $t\neq 0$ with $|t|<\min\{\varepsilon,\delta\}$. Then $q(t,1)\in A$, $q(t,2)\in B$, and $q(t,1)=q(t,2)$ by the identification. Hence $A\cap B\neq\varnothing$. So the line with two origins is locally homeomorphic to $\mathbb R$ at every point, but it is not Hausdorff. Therefore it fails the manifold axioms and is not a topological manifold. [/example] ### Countable Control The remaining global pathology is countability: a space may be locally Euclidean but too large or too scattered to admit the usual global tools of topology. The line with two origins shows why separation is needed, while second countability rules out this different failure by requiring a countable supply of open sets from which all others are built. [definition: Second Countable Space] A topological space $X$ is second countable if there is a countable collection $\mathcal B$ of open subsets of $X$ such that every open set $O\subset X$ can be written as a union of elements of $\mathcal B$. [/definition] Second countability is the bridge from local coordinates to global manageability. It suggests a concrete question: can the many chart domains in an arbitrary atlas be reduced to countably many without losing coverage of the manifold? The next theorem gives the answer that makes manifold arguments countable from the start. [quotetheorem:9911] After this countable reduction, it is natural to ask whether the topology can be described by distances, even though no metric was included in the definition. The countability theorem makes the axiom operational rather than decorative, and it lets arguments reduce from arbitrary local data to countable local data. This question leads to metrizability for ordinary geometric arguments. Although a manifold is defined topologically, the standard axioms force enough regularity that a compatible metric exists. [quotetheorem:9912] This result justifies the common habit of reasoning with sequences, distances, and sequential convergence on manifolds, while remembering that the metric is not part of the definition. Compactness itself is already a topological notion; the metric gives a convenient language for approaching it through sequences and estimates. The same hypotheses also give two other forms of global discipline that are used constantly but are easy to take for granted. Manifolds are locally compact because their chart neighbourhoods look like open subsets of Euclidean space, and second countable locally Euclidean Hausdorff spaces are paracompact. Paracompactness is what later permits partitions of unity on smooth manifolds, turning local constructions into global ones without requiring compactness. ## Boundaries and Nonexamples ### Half-Space Local Models The definition above describes manifolds without boundary. Many geometric spaces have edges, and they require a modified local model. The distinction matters because a half-space point does not have the same local topology as an interior Euclidean point. [definition: Topological Manifold with Boundary] A topological manifold with boundary of dimension $n\geq 1$ is a Hausdorff, second countable topological space $M$ such that for every point $p\in M$ there exist an open neighbourhood $U\subset M$, a subset $W\subset\mathbb H^n$ open in the subspace topology, and a homeomorphism \begin{align*} \varphi:U&\to W, \end{align*} where \begin{align*} \mathbb H^n=\{x\in\mathbb R^n:x_n\geq 0\}. \end{align*} [/definition] The half-space model creates a new coordinate test: a point may land on the hyperplane $x_n=0$ in one boundary chart or in the region $x_n>0$. For the word "boundary" to have any meaning, that test must be independent of the chosen chart. Otherwise the edge of the space would be a feature of notation rather than a feature of the space. The next theorem is the boundary analogue of invariance of dimension, and it is what makes the later definition of $\partial M$ possible. [quotetheorem:9913] With this independence in place, the half-space condition can be turned into an actual subset of $M$. This definition is needed because the original data only says that every point has some half-space chart; it does not name which points are edge points and which points are ordinary interior points. The boundary records exactly the points that every compatible coordinate system must place on the hyperplane $x_n=0$. [definition: Boundary of a Manifold with Boundary] Let $M$ be a topological manifold with boundary of dimension $n\geq 1$. The boundary of $M$ is \begin{align*} \partial M=\{p\in M:\varphi(p)_n=0\text{ for some boundary chart }(U,\varphi)\text{ with }p\in U\}. \end{align*} [/definition] Once the boundary points have been isolated, the next useful question is where the original boundaryless theory still applies without modification. A manifold with boundary contains both edge points and ordinary Euclidean points, and many arguments need to restrict to the latter before using open-set charts in $\mathbb R^n$. The interior names precisely that region, so that statements can distinguish between phenomena caused by the half-space edge and phenomena inherited from ordinary manifolds. [definition: Interior of a Manifold with Boundary] Let $M$ be a topological manifold with boundary of dimension $n\geq 1$. The interior of $M$ is \begin{align*} M^\circ=M\setminus\partial M. \end{align*} [/definition] The definition of $M^\circ$ turns the boundary construction into a clean division of the space: points of $M^\circ$ are handled by the boundaryless local theory, while points of $\partial M$ require half-space coordinates. This separation is not cosmetic; the same topological space may fail to be a manifold without boundary exactly because a few points have half-space rather than full-space neighbourhoods. [example: Closed Interval] The closed interval $[0,1]$ is not a one-dimensional topological manifold without boundary. Indeed, any open neighbourhood $U\subset[0,1]$ of $0$ contains a set of the form $[0,\varepsilon)$, because $U=[0,1]\cap G$ for some open $G\subset\mathbb R$ with $0\in G$. The connected component of $0$ in $U$ therefore has $0$ as an endpoint, so deleting $0$ leaves only the one-sided part near $0$. By contrast, if an open interval $I\subset\mathbb R$ contains a point $s$, then \begin{align*} I\setminus\{s\}=(I\cap(-\infty,s))\cup(I\cap(s,\infty), \end{align*} and both terms are nonempty open intervals for $s$ in the interior of $I$. Thus deleting the corresponding point from an interval neighbourhood produces two local sides, while deleting $0$ from a sufficiently small neighbourhood in $[0,1]$ produces only one. A homeomorphism preserves this local component structure, so no neighbourhood of $0$ is homeomorphic to an open subset of $\mathbb R$. It is, however, a one-dimensional manifold with boundary. For $0<a<1$, choose $\varepsilon>0$ with $a-\varepsilon>0$ and $a+\varepsilon<1$; then $(a-\varepsilon,a+\varepsilon)\subset[0,1]$ is carried homeomorphically to $(-\varepsilon,\varepsilon)$ by $x\mapsto x-a$. At the left endpoint, $[0,\varepsilon)$ is open in $[0,1]$ and the identity map sends it homeomorphically to $[0,\varepsilon)$, which is open in $\mathbb H^1=[0,\infty)$. At the right endpoint, $(1-\varepsilon,1]$ is open in $[0,1]$ and the map $x\mapsto 1-x$ sends it homeomorphically to $[0,\varepsilon)$ in $\mathbb H^1$. Thus the endpoints are boundary points, and the interior points have ordinary one-dimensional Euclidean charts. [/example] ### Singular Quotients Boundaries are not the only way local Euclidean behaviour can fail. Singularities can be subtler, especially when a space looks almost manifold-like away from a small set. [example: A Cone That Is Only a Manifold with Boundary] Let $C$ be the quotient of $S^1\times[0,1]$ obtained by identifying all points of $S^1\times\{0\}$ to one point, and let $q:S^1\times[0,1]\to C$ be the quotient map. Define \begin{align*} F:S^1\times[0,1]\to\mathbb R^2,\qquad F(u,r)=ru. \end{align*} If $u=(\cos\theta,\sin\theta)$, then \begin{align*} F(u,r)=(r\cos\theta,r\sin\theta). \end{align*} Thus \begin{align*} \|F(u,r)\|^2=(r\cos\theta)^2+(r\sin\theta)^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2(\cos^2\theta+\sin^2\theta)=r^2\leq 1, \end{align*} so $F$ lands in the closed disk $D^2=\{z\in\mathbb R^2:\|z\|\leq 1\}$. Also $F(u,0)=0$ for every $u\in S^1$, so $F$ is constant on the collapsed circle and therefore induces a map \begin{align*} \overline F:C\to D^2,\qquad \overline F(q(u,r))=ru. \end{align*} This induced map is bijective. If $z\in D^2$ and $z\neq 0$, then $z=\|z\|(z/\|z\|)$ with $z/\|z\|\in S^1$ and $0<\|z\|\leq 1$, so $z=\overline F(q(z/\|z\|,\|z\|))$. The point $0\in D^2$ is $\overline F(q(u,0))$ for every $u\in S^1$. Conversely, if $ru=sv$ with $u,v\in S^1$ and $r,s\in[0,1]$, then taking norms gives \begin{align*} r=\|ru\|=\|sv\|=s. \end{align*} If $r=s>0$, dividing $ru=rv$ by $r$ gives $u=v$, so $q(u,r)=q(v,s)$. If $r=s=0$, both points lie in the collapsed circle and again have the same image in $C$. Since $S^1\times[0,1]$ is compact and $D^2$ is Hausdorff, the induced continuous bijection $\overline F:C\to D^2$ is a homeomorphism by the *compact-to-Hausdorff homeomorphism criterion*. Under this homeomorphism, the collapsed circle $S^1\times\{0\}$ becomes the centre $0\in D^2$, not a singular point. For instance, the inverse image of a small open disk $B(0,\varepsilon)\subset D^2$ is a neighbourhood of the collapsed point, and $B(0,\varepsilon)$ is homeomorphic to an open subset of $\mathbb R^2$ by the identity map. The points coming from $S^1\times\{1\}$ become the boundary circle $\partial D^2$. Near $(1,0)\in\partial D^2$, use the coordinates \begin{align*} (u,v)=(1-x,y). \end{align*} The disk condition $x^2+y^2\leq 1$ becomes \begin{align*} (1-u)^2+v^2\leq 1. \end{align*} Expanding gives \begin{align*} 1-2u+u^2+v^2\leq 1. \end{align*} Subtracting $1$ gives \begin{align*} u^2+v^2\leq 2u, \end{align*} so in a sufficiently small neighbourhood of $(1,0)$ the allowed points lie on one side of the tangent line $u=0$. These are half-disk neighbourhoods, matching the local model for a two-dimensional manifold with boundary, not ordinary disk neighbourhoods in $\mathbb R^2$. Thus the cone is a two-dimensional manifold with boundary, and the boundary is the image of $S^1\times\{1\}$. The drawing may suggest that the collapsed tip is the dangerous point, but the actual obstruction to being a manifold without boundary is the outer boundary circle. [/example] This example warns against trusting pictures alone. A space may be smooth-looking almost everywhere but fail the manifold condition at a small collection of points. ## Coordinate Changes ### Transition Maps Even before differentiability enters, overlapping charts must agree topologically. The map from one coordinate description to another is the transition map. It is the place where extra structures are later imposed. [definition: Transition Map] Let $(U,\varphi)$ and $(V,\psi)$ be coordinate charts on a topological manifold $M$. If $U\cap V\neq\varnothing$, the transition map from $\varphi$-coordinates to $\psi$-coordinates is \begin{align*} \psi\circ\varphi^{-1}:\varphi(U\cap V)&\to\psi(U\cap V). \end{align*} [/definition] Because $\varphi$ and $\psi$ are homeomorphisms, every transition map for a topological manifold is a homeomorphism between open subsets of $\mathbb R^n$. This is exactly enough to transfer open sets, connectedness, compactness, and other topological properties between coordinate patches. The following theorem records that compatibility as a reusable fact, because later structures are imposed by strengthening precisely this transition-map condition. [quotetheorem:9914] This theorem is simple but important. It identifies the exact interface between topology and later geometry. A smooth manifold asks these transition maps to be smooth; a complex manifold asks them to be holomorphic; a piecewise-linear manifold asks for a different kind of compatibility. ### A Concrete Change of Coordinates [example: Two Charts on the Circle] Let $N=(0,1)$ and $S=(0,-1)$, and set $U_N=S^1\setminus\{N\}$ and $U_S=S^1\setminus\{S\}$. Both are open in $S^1$ because singletons are closed in the subspace topology inherited from $\mathbb R^2$. Use stereographic projection from $N$ and from $S$ to the horizontal axis: \begin{align*}\varphi_N(x,y)=\frac{x}{1-y}\end{align*} for $(x,y)\in U_N$, and \begin{align*}\varphi_S(x,y)=\frac{x}{1+y}\end{align*} for $(x,y)\in U_S$. The inverse formulas are \begin{align*}\varphi_N^{-1}(t)=\left(\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right)\end{align*} and \begin{align*}\varphi_S^{-1}(u)=\left(\frac{2u}{u^2+1},\frac{1-u^2}{u^2+1}\right).\end{align*} For example, \begin{align*}\left(\frac{2t}{t^2+1}\right)^2+\left(\frac{t^2-1}{t^2+1}\right)^2=\frac{4t^2+(t^2-1)^2}{(t^2+1)^2}\end{align*} and \begin{align*}4t^2+(t^2-1)^2=4t^2+t^4-2t^2+1=t^4+2t^2+1=(t^2+1)^2,\end{align*} so $\varphi_N^{-1}(t)\in S^1$; the verification for $\varphi_S^{-1}$ is the same with the second coordinate sign reversed. Since these formulas are rational functions with denominator $t^2+1>0$, the charts and their inverses are continuous. The overlap is \begin{align*}U_N\cap U_S=S^1\setminus\{N,S\}.\end{align*} In $\varphi_N$-coordinates, the missing south pole is \begin{align*}\varphi_N(0,-1)=\frac{0}{1-(-1)}=0,\end{align*} so $\varphi_N(U_N\cap U_S)=\mathbb R\setminus\{0\}$. If $t\neq 0$, then \begin{align*}(\varphi_S\circ\varphi_N^{-1})(t)=\varphi_S\left(\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right).\end{align*} Substituting into the formula for $\varphi_S$ gives \begin{align*}(\varphi_S\circ\varphi_N^{-1})(t)=\frac{2t/(t^2+1)}{1+(t^2-1)/(t^2+1)}.\end{align*} The denominator is \begin{align*}1+\frac{t^2-1}{t^2+1}=\frac{t^2+1+t^2-1}{t^2+1}=\frac{2t^2}{t^2+1},\end{align*} hence \begin{align*}(\varphi_S\circ\varphi_N^{-1})(t)=\frac{2t/(t^2+1)}{2t^2/(t^2+1)}=\frac{1}{t}.\end{align*} The map $t\mapsto 1/t$ sends $\mathbb R\setminus\{0\}$ to itself, is continuous there, and is its own inverse because \begin{align*}\frac{1}{1/t}=t\end{align*} for every $t\neq 0$. Thus the two stereographic charts cover the circle, and their coordinate change on the overlap is a homeomorphism. [/example] Coordinate changes also explain why local statements must be invariant under homeomorphism. A property that depends on a specific coordinate formula is not a topological property unless it survives every allowed transition map. ## Compactness, Connectedness, and Global Shape ### Connected Components Once a space is known to be a manifold, local Euclidean facts combine with global topology. Compactness and connectedness are not part of the definition, but they determine much of the shape of examples. [definition: Connected Manifold] A connected manifold is a topological manifold whose underlying topological space is connected. [/definition] Connectedness prevents the manifold from decomposing into disjoint open pieces. A disconnected manifold is still a manifold, but many classification statements are naturally made component by component. Since Euclidean neighbourhoods are locally connected, the components of a manifold should not be invisible or densely interwoven; the next theorem makes that expectation precise. [quotetheorem:1003] After separating a manifold into components, the next global question is whether each component can be controlled by finitely many coordinate charts. The openness of components reflects the local connectedness of Euclidean open sets: components are not hidden fractal pieces, but open regions in the manifold topology. ### Compact Manifolds This need leads to compactness, which gives a different kind of global control. A compact manifold can be covered by finitely many coordinate charts, which is often the first step in reducing a global argument to finitely many Euclidean arguments. [quotetheorem:9915] This is one of the main practical reasons compact manifolds are easier to handle than noncompact ones. Finitely many charts mean finitely many local estimates, finitely many transition regions, and no need to control behaviour at infinity. [example: Compact and Noncompact Surfaces] For $S^2=\{(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=1\}$, stereographic projection from the north and south poles gives two coordinate charts onto open subsets of $\mathbb R^2$, so $S^2$ is a two-dimensional manifold. It is compact because it is closed and bounded in $\mathbb R^3$: boundedness follows from $\|(x,y,z)\|=1$, and closedness follows since $S^2=f^{-1}(\{1\})$ for the [continuous function](/page/Continuous%20Function) \begin{align*} f(x,y,z)=x^2+y^2+z^2. \end{align*} The torus $T^2$ can be represented as the quotient of $[0,1]^2$ by identifying opposite edges. The square $[0,1]^2$ is compact, and the quotient map sends it continuously onto $T^2$, so $T^2$ is compact. The local charts are the square charts away from the edges, edge-crossing charts near identified edges, and corner charts near the four identified corners; each has image an open subset of $\mathbb R^2$. Hence $T^2$ is also a compact two-dimensional manifold. The plane $\mathbb R^2$ is a two-dimensional manifold by the identity chart \begin{align*} \operatorname{id}_{\mathbb R^2}:\mathbb R^2\to\mathbb R^2. \end{align*} It is not compact: the open sets $B(0,k)$ for $k\in\mathbb N$ cover $\mathbb R^2$, since every $x\in\mathbb R^2$ has $\|x\|<k$ for some integer $k$, but no finite subcollection covers $\mathbb R^2$. Indeed, if $B(0,k_1),\ldots,B(0,k_m)$ are chosen and $K=\max\{k_1,\ldots,k_m\}$, then the point $(K+1,0)$ is not in any of them because \begin{align*} \|(K+1,0)\|=K+1>K\geq k_i \end{align*} for every $i$. The open cylinder $S^1\times\mathbb R$ is a two-dimensional manifold: near $(p,a)$, choose a chart on $S^1$ around $p$ with coordinate $s$, and combine it with the interval coordinate $t-a$ on $(a-\varepsilon,a+\varepsilon)\subset\mathbb R$. It is not compact, because the open sets \begin{align*} S^1\times(-k,k) \end{align*} for $k\in\mathbb N$ cover $S^1\times\mathbb R$, but no finite subcollection covers it. If $K$ is the largest chosen integer, then for any $p\in S^1$ the point $(p,K+1)$ is outside every chosen set. Thus all four spaces are locally modeled on $\mathbb R^2$, while compactness separates the closed surfaces $S^2$ and $T^2$ from $\mathbb R^2$ and $S^1\times\mathbb R$. [/example] The local definition therefore leaves room for substantial global variety. Dimension fixes the local model, while compactness, connectedness, orientability, and fundamental group record how the pieces fit together. ## Beyond and Connected Topics Topological manifolds are the entry point for several more structured theories. A [smooth manifold](/page/Smooth%20Manifold) adds differentiable compatibility to the transition maps, which allows tangent vectors, vector fields, differential forms, and integration. The topological definition remains underneath: every smooth manifold is first a topological manifold with a special kind of atlas. Riemannian geometry adds a smoothly varying [inner product](/page/Inner%20Product) on tangent spaces. This turns local coordinates into a setting for length, angle, geodesics, curvature, and volume. Androma's [Cambridge III Riemannian Geometry](/page/Cambridge%20III%20Riemannian%20Geometry) is a natural continuation after the smooth theory is in place. Differential topology studies which topological manifolds admit smooth structures and how maps between manifolds behave up to deformation. In low dimensions, topology and smoothness interact in especially delicate ways; in high dimensions, surgery and cobordism become central tools. Algebraic topology extracts invariants such as homology, cohomology, and fundamental groups from manifolds. Androma's [Differential Forms II: Manifolds and Cohomology](/page/Differential%20Forms%20II%3A%20Manifolds%20and%20Cohomology) connects the local coordinate picture to cohomological invariants through differential forms. Algebraic geometry uses spaces that can locally resemble zero sets of polynomials rather than open subsets of Euclidean space. The comparison with manifolds clarifies why singular points require special treatment. Androma's [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry) gives one route into that broader geometric setting. ## References Androma, [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry). Androma, [Smooth Manifold](/page/Smooth%20Manifold). Androma, [Differential Forms II: Manifolds and Cohomology](/page/Differential%20Forms%20II%3A%20Manifolds%20and%20Cohomology). Androma, [Cambridge III Riemannian Geometry](/page/Cambridge%20III%20Riemannian%20Geometry). Androma, [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). John M. Lee, *Introduction to Topological Manifolds* (2011). James R. Munkres, *Topology* (2000). Allen Hatcher, *Algebraic Topology* (2002).