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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.

The codimension is the difference between the dimension of a manifold and that of its submanifold. For complex submanifolds, complex codimension one means real codimension two.

A Lie group is a group that is also a finite-dimensional smooth manifold, with multiplication $G\times G\to G$ and inversion $G\to G$ both smooth maps. Its tangent space at the identity element carries a natural Lie algebra structure. Smoothness lets us differentiate the group operation; being a group means that every transformation can be composed and undone. The dimension of a manifold counts independent local parameters, rather than the number of elements of the group. See MIT 18.745, Lecture 2, Definition 2.1 and Example 2.5.

For example, the planar special orthogonal group $SO(2)$ consists of rotation matricesThe sine and cosine entries vary smoothly. Four entries appear, but they are constrained by $R_{\theta }^T{R}_{\theta }=I$ and $\det R_{\theta }=1$: locally, one angle specifies the whole matrix. Multiplication adds angles, with angles differing by $2\pi$ representing the same rotation. Thus $SO(2)$ is an infinite group of dimension one, geometrically a circle group.

The positive-dimensional examples motivate the language of continuous symmetry, but the definition also admits finite groups with the discrete topology as zero-dimensional Lie groups. Conversely, being an infinite group alone says nothing about having a continuous angle parameter. The compatibility between smooth maps and the group operation is what makes differentiation useful.

The Arnold-Liouville theorem concerns a connected compact component $L$ of a regular common level of $n$ commuting independent smooth functions $F=(F_1,\dots ,F_n)$ on a $2n$-dimensional symplectic manifold. It states that $L$ is a Lagrangian torus, and that a neighbourhood of $L$ has action-angle variablesin which every $F_i$ depends only on $I$. If $H=F_1$ or, more generally, $H=h(F)$, its Hamiltonian flow satisfiesAngles have period $2\pi$ here. The theorem is local near this regular compact component; it does not assert globally defined action-angle variables across singular levels or over an entire base with monodromy.

First, $L$ has dimension of a manifold $n$ by the submersion theorem. The Hamiltonian vector fields $X_{F_i}$ are independent, tangent to $L$, and commute. They span $TL$. Moreoverso $L$ is a Lagrangian submanifold. Compactness makes these restricted vector fields complete. Their joint flow is an ${\mathbb{R}}^n$-action on $L$. Its orbits are open because the fields span $TL$, so connectedness gives one orbit. The stabilizer $\Lambda$ of a point is discrete by the inverse function theorem, andCompactness forces $\Lambda$ to be a full-rank Euclidean lattice: otherwise an unbounded linear coordinate transverse to its span would descend to the quotient. Thus $L$ is an $n$-torus.

Next choose a small ball of regular values near $F(L)$ and a neighbourhood $N$ of this component which is a product family of compact tori $L_c$. This local trivialization follows directly by choosing transverse vector fields $Z_j$ with $d{F}_i(Z_j)={\delta }_{ij}$ and lifting short radial paths in the base; compactness of $L$ gives a uniform neighbourhood where their flows exist. Hence $N$ deformation retracts onto $L$. The commuting joint flows on nearby fibres have smoothly varying full-rank Hamiltonian period lattices. A basis $t^{(j)}(c)\in {\mathbb{R}}^n$ of their periods can be chosen smoothly on this small ball: continue the return equations from a basis at the central fibre, using their nonsingular vertical flow derivatives and the implicit function theorem. No global choice of lattice basis is needed.

Because $\omega {|}_L=0$ and $N$ retracts onto $L$, its closed symplectic form is exact on $N$. Choose a one-form $\lambda$ with $d\lambda =-\omega$. Let ${\gamma }_j(c)$ be the smoothly continued cycles represented by the period basis and define the action integralsFor a transverse variation $v$ of the fibre, differentiating a cycle integral and using Cartan's magic formula eliminates the integral of the exact term. With the cycle parametrized by the joint-flow time $t^{(j)}$, this givesThusThe period matrix is nonsingular, so $I$ gives coordinates on the base. The corresponding Hamiltonian vector fields satisfyTheir time-$2\pi$ flows return along the basis cycles. They commute because each $I_j$ is a function of the commuting $F_i$. Consequently they give a free torus action on $N$.

Choose a section over the action-coordinate ball and use this torus action to define angles $\theta$. Since $X_{I_j}={\partial }_{{\theta }_j}$, the symplectic form takes the formThe closed base two-form $\beta$ is exact on the ball: write $\beta =d\alpha$, $\alpha ={\sum }_j{a}_j(I)d{I}_j$, by the Poincaré lemma. Changing the angular origins by ${\theta }_j^{\prime }={\theta }_j+a_j(I)$ removes this term, since ${\sum }_{j}d{\theta }_j^{\prime }\wedge d{I}_j=\omega$. This completes the construction of action-angle variables. The functions $F_i$ are constant on the fibres, so are functions of $I$ alone, and Hamilton's equations give the claimed straight-line motion. This proves both the topological and symplectic parts of the Arnold-Liouville theorem.

Let $o=(0,0)$ be the common point of the two circles. For sufficiently small $\epsilon >0$, the open neighbourhoodhas the following local structure: $W\setminus \{o\}$ consists of four disjoint open arcs, an upper and a lower branch from each circle. These are its four connected components, each open and closed relative to $W\setminus \{o\}$.

Suppose a coordinate chart of some dimension of a manifold $n$ existed at $o$. The case $n=0$ would make $o$ an isolated point, which it is not. For $n\ge 1$, shrink the chart inside $W$ so that its image is an open ball centred at the image of $o$. Its domain of a function $V$ is an open neighbourhood of $o$, so it meets all four arms of $W$. Intersecting those arms with $V\setminus \{o\}$ gives four disjoint nonempty clopen sets partitioning $V\setminus \{o\}$; hence there are at least four connected components.

But a punctured ball in $\mathbb{R}$ has exactly two connected components, and a punctured ball in ${\mathbb{R}}^n$ for $n\ge 2$ is path-connected. A homeomorphism preserves connected components, giving a contradiction in every dimension of a manifold. Thus $X$ is not a topological manifold. The argument uses arbitrarily small coordinate charts; the fact that the whole $X\setminus \{o\}$ has two pieces alone would not be a local obstruction.

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]

A topological tripod is three copies of the closed interval $[0,1]$ with their zero endpoints identified to a single junction $o$. Equivalently, embed three line segments with distinct directions in the Euclidean plane and use the subspace topology. It is a compact, connected metric space, and is Hausdorff, second countable, and paracompact. Every interior arm point has a coordinate chart to an open interval, and each outer endpoint has a half-interval chart as in a topological manifold with boundary. The junction has neither kind of chart.

Here is a precise local obstruction. Let $S$ be a small open neighbourhood of $o$ consisting of initial segments of the three arms, excluding their outer endpoints. Every connected open neighbourhood $N$ of $o$ contained in $S$ contains an initial segment on every arm, because $N$ is open at $o$. Its intersection with each arm has no gaps: if a point between $o$ and a point of $N$ were missing, cutting there would separate the part beyond it from the rest of $N$, contradicting connectedness. Thus $N\setminus \{o\}$ has exactly three connected components, one interval per arm.

Suppose a coordinate chart $h$ to the real line existed at $o$. Choose a small open interval $I$ about $h(o)$ lying inside $h(S\cap \mathrm{domain}(h))$, and restrict the chart to $N=h^{-1}(I)$. This is a connected open neighbourhood of $o$ contained in $S$. Removing $o$ leaves three connected components in $N$, while removing $h(o)$ leaves two in $I$. A homeomorphism preserves connected components, so this is impossible. This restriction step is essential: the count is proved for any sufficiently small connected chart neighborhood, not merely one chosen drawing.

A higher dimension of a manifold cannot repair the junction. For $n\ge 2$, repeat the argument with an open ball about $h(o)$: a punctured open ball in ${\mathbb{R}}^n$ is path-connected, whereas $N\setminus \{o\}$ still has three connected components. A zero-dimensional chart would make $o$ an isolated point, which it is not. A chart at a boundary point of a topological manifold with boundary similarly gives a punctured relative half-ball: one connected component when $n=1$, and a path-connected space when $n\ge 2$. Consequently the topological tripod fails the local definition even when intrinsic boundary is allowed.

The circle provides a useful warning about scale: deleting a point from the whole circle leaves a connected space, but deleting it from a sufficiently small connected arc leaves two connected components. Local chart tests use the latter. The circle and interval are not homeomorphic article uses deletion from the whole spaces for a different, global obstruction.