A point on a circle has no preferred real coordinate. A small arc does. A manifold records such local coordinates and requires their changes to be smooth.
2.1. Local coordinates
Definition 2.1. Topological manifold and chart
A topological \(n\)-manifold without boundary is a Hausdorff, second-countable topological space \(M\) in which each point has an open neighbourhood homeomorphic to an open subset of \(\mathbb R^n\). Hausdorff means distinct points have disjoint open neighbourhoods. Second-countable means there is a countable family of open sets such that every open set is a union of members of that family.
A chart is a pair \((U,x)\) with \(U\subset M\) open and \(x:U\to x(U)\subset\mathbb R^n\) a homeomorphism onto an open set. A homeomorphism is a continuous bijection with continuous inverse. The functions \(x^i=\pi_i\circ x\) are the coordinate functions; \(\pi_i\) selects the \(i\)-th entry.
The local Euclidean condition supplies coordinates. Hausdorffness separates distinct points. Second-countability rules out spaces requiring uncountably many separate coordinate regions. These are conventions restricting the class of spaces; they are not consequences of local coordinates. “Without boundary” specifies open Euclidean images, rather than half-space images.
Definition 2.2. Compatibility and atlas
On overlapping charts \((U,x),(V,y)\), the transition map is \(y\circ x^{-1}:x(U\cap V)\to y(U\cap V)\). The charts are smoothly compatible if this map and its inverse are smooth; disjoint charts are compatible by convention. A smooth atlas is a family of pairwise compatible charts whose domains cover \(M\). A maximal smooth atlas contains every chart compatible with all its charts. A smooth manifold is a topological manifold equipped with a maximal smooth atlas, its smooth structure. A chart belonging to this maximal atlas is called a smooth chart.
The transition images are open because charts are homeomorphisms. Requiring smooth transitions makes differentiation agree when two coordinates describe the same region.
Proposition 2.3. An atlas determines a unique smooth structure
Every smooth atlas lies in a unique maximal smooth atlas.
Proof.
Add every chart compatible with the given atlas \(\mathcal A\). Two added charts \(x,y\) have a smooth transition near each overlap point: choose an original chart \(z\) through that point and factor \(y\circ x^{-1}=(y\circ z^{-1})\circ(z\circ x^{-1})\). The chain rule proves smoothness, locally at every point, in both directions. Thus the enlarged family is an atlas and is maximal. Any maximal atlas containing \(\mathcal A\) is contained in this enlargement; all charts of the enlargement are compatible with that maximal atlas by the same factorization, so maximality gives equality.
□Example. The identity chart on \(\mathbb R^n\) determines its standard smooth structure. The chart \(y=2x\) belongs to it because its transitions are linear and smooth. On the topological line, the single chart \(y=x^3\) also generates an atlas, but it is incompatible with the identity chart: the inverse transition is not differentiable at zero. An atlas chooses differentiability, not just topology.
2.2. The circle, with no missing point
Let \(S^1=\{(X,Y):X^2+Y^2=1\}\) with its subspace topology. Remove \(N=(0,1)\) for one chart and \(S=(0,-1)\) for the other:
\[u=\frac X{1-Y},\qquad v=\frac X{1+Y}.\]The inverse of the first is \(u\mapsto(2u/(1+u^2),(u^2-1)/(1+u^2))\); the inverse of the second is \(v\mapsto(2v/(1+v^2),(1-v^2)/(1+v^2))\). Substitution verifies both inverse identities. All displayed maps are continuous, since their denominators never vanish on their domains. Both chart images are \(\mathbb R\). On the overlap, \(uv=X^2/(1-Y^2)=1\), so the transition is \(v=1/u\), smooth on \(\mathbb R\setminus\{0\}\) with smooth inverse.
The circle is Hausdorff as a subspace of a metric space. Intersecting it with Euclidean balls having rational centres and positive rational radii gives a countable basis: every open ball about a point contains such a smaller ball containing that point. Thus these charts make it a smooth one-manifold.
Definition 2.4. Smooth maps between manifolds
A continuous map \(F:M\to N\) is \(C^k\) if every coordinate representative \(y\circ F\circ x^{-1}\), restricted to \(x(U\cap F^{-1}(V))\), is \(C^k\). It is smooth if it is \(C^\infty\). A manifold diffeomorphism is a smooth bijection with smooth inverse. An open subset inherits charts by restriction.
Proposition 2.5. One atlas is enough
Smoothness can be checked in covering atlases. Identities and composites of smooth maps are smooth.
Proof.
Any other representative is the tested one composed on each side with smooth transitions. The chain rule proves the assertion locally, hence everywhere. Identity representatives are transitions. A composite representative is a composite of representatives, after choosing a chart about the intermediate point.
□Example. The height \(H:S^1\to\mathbb R\), \(H(X,Y)=Y\), is smooth: in the two charts its representatives are \((u^2-1)/(1+u^2)\) and \((1-v^2)/(1+v^2)\). These are quotients of polynomials with positive denominators. The inclusion \(S^1\to\mathbb R^2\) is smooth by the inverse-chart formulas above.
2.3. Constraints give manifolds
Definition 2.6. Embedded submanifold and regular value
A subset \(Q\) of an \(m\)-manifold is an embedded \(d\)-submanifold if it has the subspace topology and, near every point of \(Q\), an ambient smooth chart takes \(Q\) to the intersection of the chart image with \(\mathbb R^d\times\{0\}\). Restricted first coordinates are its charts.
For a smooth \(G:O\subset\mathbb R^m\to\mathbb R^r\), a point \(a\) is regular if \(DG_a\) is surjective, equivalently has rank \(r\). A value \(b\) is regular if every \(a\in G^{-1}(b)\) is regular. Rank means the dimension of the linear image.
Proposition 2.7. Regular-level theorem
If \(b\) is a regular value, then the nonempty level set \(Q=G^{-1}(b)\) is an embedded \((m-r)\)-submanifold.
Proof.
At \(a\in Q\), select \(r\) columns of \(DG_a\) forming an invertible minor and reorder coordinates as \((u,v)\in\mathbb R^{m-r}\times\mathbb R^r\). The map \(H(u,v)=(u,G(u,v)-b)\) has invertible derivative: its block matrix has identity in the upper left, zero in the upper right, and the chosen invertible minor in the lower right. The inverse function theorem, proved below, makes \(H\) a local diffeomorphism. It takes the level set exactly to the last-\(r\)-coordinates-zero slice. Restricted charts have smooth transitions because ambient ones do. Hausdorffness and a countable basis descend from \(\mathbb R^m\) by intersection with the subset.
□Example. For a pendulum of unit length, \(G(X,Y)=X^2+Y^2\), and \(DG=2(X,Y)\) has rank one on \(G^{-1}(1)\). There is one local coordinate, not two independent ones. For \(G(X,Y)=XY\), the value zero fails regularity at the origin. Its level set is a crossing of two lines; deleting the origin from any sufficiently small connected neighbourhood leaves four branches, whereas deleting a point from an interval leaves two. It is not a one-manifold there.
Proof of the inverse function theorem used above
Statement. If \(F\) is \(C^1\) near \(a\) and \(DF_a\) is invertible, then \(F\) restricts to a \(C^1\) diffeomorphism between neighbourhoods of \(a\) and \(F(a)\). If \(F\) is smooth, its local inverse is smooth. This is the smooth inverse function theorem.
Proof. Translations and multiplication by \((DF_a)^{-1}\) reduce to \(a=F(a)=0\), \(DF_0=I\). Choose a closed ball \(\overline B_r\) inside the domain with \(\|DF_x-I\|\le q<1\). The fundamental theorem on a segment gives
\[|(F(x)-x)-(F(z)-z)|\le q|x-z|.\]For \(|y|<(1-q)r\), the map \(T_y(x)=y+x-F(x)\) takes \(\overline B_r\) into itself and contracts distances by \(q\). Iterating from zero gives \(|x_{j+1}-x_j|\le q^j|x_1-x_0|\). Summing the geometric tail makes the sequence Cauchy, so it converges in the closed Euclidean ball. Continuity gives a fixed point; two fixed points have distance at most \(q\) times that distance, so coincide. Denote it by \(K(y)\). Its norm is at most \(|y|/(1-q)<r\). The same inequality for two right-hand sides gives \(|K(y)-K(z)|\le|y-z|/(1-q)\).
Thus \(F\) is a homeomorphism from \(B_r\cap F^{-1}(B_{(1-q)r})\) onto \(B_{(1-q)r}\). For \(x=K(y)\), set \(\Delta x=K(y+h)-K(y)=O(|h|)\). Expansion gives \(h=DF_x\Delta x+o(|h|)\), hence \(DK_y=(DF_{K(y)})^{-1}\). The inverse matrix is continuous, and smooth wherever its determinant is nonzero, because its entries are cofactors divided by the determinant. The derivative formula proves \(K\in C^1\), then inductively \(K\in C^k\) whenever \(F\in C^k\). This proves the statement.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. Two charts and their maximal extension
Applies to: Definition 2.1, Definition 2.2, Proposition 2.3.
On the topological line take \(x(p)=p\) and \(y(p)=2p+1\), both with domain \(\mathbb R\). The space is Hausdorff (disjoint short intervals separate distinct points) and second-countable (intervals with rational endpoints are a countable basis). Both chart images are open, and the maps and inverses are continuous. Their transitions \(y=2x+1\), \(x=(y-1)/2\) are smooth, so they form a smooth atlas.
A further chart \(z(p)=p+p^3\) belongs to its maximal atlas: its derivative \(1+3p^2\) never vanishes, it is strictly increasing onto \(\mathbb R\), and the inverse-function theorem gives a smooth inverse near every point. A chart compatible with the identity is compatible with \(y\) by composition with an affine map; conversely any chart compatible with this atlas is compatible with the identity. Thus adjoining all such charts is exactly its unique maximal extension. The proposed chart \(w(p)=p^3\) does not belong: its inverse has no derivative at zero. This illustrates both compatibility and the actual smooth structure selected by the atlas.
Further proof details
The inverse of \(z=p+p^3\) can also be exhibited, so its smoothness need not rest on invoking the inverse-function theorem. Define \(s=\tfrac13\operatorname{arsinh}(3\sqrt3 z/2)\) and \(p=2\sinh(s)/\sqrt3\), where \(\operatorname{arsinh}u=\log(u+\sqrt{1+u^2})\), smooth for all real \(u\) because its logarithm argument is positive. With \(\sinh s=(e^s-e^{-s})/2\), expansion gives \(\sinh3s=3\sinh s+4\sinh^3s\); consequently \(p+p^3=2\sinh3s/(3\sqrt3)=z\). The map \(p\mapsto p+p^3\) is strictly increasing, since for \(p>q\) the difference is \((p-q)(1+p^2+pq+q^2)>0\); the quadratic term equals \((p+q/2)^2+3q^2/4\). Thus this right inverse is its unique inverse and both directions are smooth. Affine transition composition then proves compatibility with the stated atlas.
Worked example 2. Checking smoothness in coordinates
Applies to: Definition 2.4, Proposition 2.5.
Use the preceding smooth line and \(F(p)=p^2\) with standard target coordinate. In the \(x\)-chart the representative is \(x^2\); in the \(y\)-chart it is \(((y-1)/2)^2\). Both are smooth by explicit polynomial formulas. For any other compatible source or target chart its representative is a composite of this one with smooth transitions, so no additional test is required. The identity has representative \(y\circ x^{-1}=2x+1\), and composing \(F\) with the smooth map \(q\mapsto q+1\) gives \(p^2+1\). Thus the atlas test, identity, and composition assertions are all realized. The map \(F\) is not a diffeomorphism since it is not injective; the affine map \(p\mapsto2p+1\) is, with the displayed smooth inverse.
Worked example 3. Flattening a regular level set explicitly
Applies to: Definition 2.6, Proposition 2.7.
Let \(G:\mathbb R^2\to\mathbb R\), \(G(x,y)=x^2+y^2\). On \(G^{-1}(1)\), \(DG=(2x,2y)\) has rank 1, so 1 is a regular value. Near \((1,0)\), use the ambient chart \(\Psi(x,y)=(u,v)=(y,x^2+y^2-1)\), restricted to a small open neighborhood with \(x>0\). Its inverse is \((u,v)\mapsto(\sqrt{1+v-u^2},u)\), smooth wherever the radicand is positive. The circle is precisely \(v=0\) there. Repeating near any circle point with a nonzero coordinate supplies such charts. Their restrictions have the subspace topology and one free coordinate, checking embeddedness and the dimension \(2-1=1\) predicted by the theorem.
Further proof details
To cover every circle point explicitly, if \(x\ne0\) restrict to its sign \(\epsilon\in\{1,-1\}\) and use \((u,v)=(y,x^2+y^2-1)\), with inverse \((\epsilon\sqrt{1+v-u^2},u)\) on a sufficiently small open image. If \(y\ne0\), interchange \(x,y\) in these formulas and choose the sign of \(y\). At least one case holds since \(x^2+y^2=1\). Each inverse is smooth on positive radicand; \(v=0\) is exactly the circle, so these charts prove embeddedness at every point, not just at \((1,0)\).
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. A singular level can still be a manifold
Related statements: Definition 2.6, Proposition 2.7.
Claim refuted. If a value is not regular, its level set cannot be an embedded manifold.
Proof.
Take \(G:\mathbb R\to\mathbb R\), \(G(x)=x^2\), and the value 0. Its sole preimage is 0, where \(DG_0=0\) is not onto; thus 0 is not a regular value. Nevertheless \(G^{-1}(0)=\{0\}\) is an embedded zero-dimensional submanifold. Indeed the identity ambient chart sends this subset to \(\mathbb R^0\times\{0\}=\{0\}\), and its subspace topology is the one-point topology. All the embedded-submanifold requirements hold directly. Regularity is a sufficient test for the theorem’s conclusion, not a characterization of every manifold level set. By contrast the original \(xy=0\) crossing shows that singularity can also produce a nonmanifold; neither outcome follows merely from a rank failure.
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