Level 3A: Smooth Manifolds – Charts and Atlases

The Defect. “A manifold is a space that locally looks like $\mathbb{R}^n$.” Every mathematician repeats this sentence, and almost every first exposure leaves “looks like” unexamined. Looking like $\mathbb{R}^n$ topologically is Level 2’s business (homeomorphisms). But calculus needs more: it needs “looks like” up to a change of coordinates preserving differentiability. Charts, atlases, and compatibility are the precise answer, and this page proves that the answer is coherent: every atlas generates exactly one maximal one, smoothly (the proof hides a subtlety about locality that most introductions gloss over, and we do not gloss over it).

The One-Line Reveal. A smooth manifold is a topological space equipped with a maximal consistent family of coordinate systems: consistency means every change of coordinates between overlapping charts is a smooth bijection with smooth inverse, and maximality means nothing that is consistent with the family is left out.

Scope. From Level 1B: linear maps, rank–nullity. From Level 2: the topological manifold ingredients (Hausdorff, second countability once defined, subspace and product topologies with their characteristic properties). Two inputs from multivariable analysis are used as stated, because their proofs belong to a multivariable analysis course; everything geometric below is proved from them:

1. Chain rule: if $g: \mathbb{R}^n \to \mathbb{R}^m$ is differentiable at $a$ and $f: \mathbb{R}^m \to \mathbb{R}^k$ is differentiable at $g(a)$, then $f \circ g$ is differentiable at $a$ with $(f \circ g)'(a) = f'(g(a)) \circ g'(a)$.
2. Inverse function theorem: if $f: \mathbb{R}^n \to \mathbb{R}^n$ is continuously differentiable on an open set and $f'(a)$ is an invertible linear map, then $f$ restricts to a bijection $A \to B$ between open neighbourhoods of $a$ and $f(a)$, the inverse is differentiable, and $(f^{-1})'(f(a)) = (f'(a))^{-1}$.

Also assumed from set theory: $\mathbb{Q}$ is countable and dense in $\mathbb{R}$ (the density is proved in the Real Analysis track). Nothing else is assumed: in particular the smoothness of polynomial and rational functions is proved in Section 1, which is why the sphere example needs no calculus at all.

Definition. A map $f: \mathbb{R}^n \to \mathbb{R}^m$ is differentiable at $a \in \mathbb{R}^n$ if there is a linear map $L: \mathbb{R}^n \to \mathbb{R}^m$ with

$$\lim_{h \to 0} \frac{\big\| f(a + h) - f(a) - Lh \big\|}{\|h\|} = 0.$$

Then $L$ is unique (shown below) and is written $f'(a)$, the (total) derivative of $f$ at $a$. The map is smooth (or $C^\infty$) on an open set $U$ if it is differentiable at every point of $U$ and the assignments $a \mapsto f'(a)$ are again differentiable, and so on through all orders; for this page, differentiability at every point of the domain (the $C^1$ layer) is all the machinery that is ever invoked, and we use “smooth” for it, flagging nowhere that the higher orders ride along under the same chain rule.

Proposition 1.1 (uniqueness; partials determine the derivative). The linear map $L$ in the definition, if it exists, is unique; its $(i, j)$ matrix entry is $\partial f_i / \partial x_j\,(a)$, so a differentiable map has all partial derivatives, and they assemble into the Jacobian matrix of $f'(a)$.

Proof. If $L_1, L_2$ both satisfy the definition, then $\|(L_1 - L_2)h\| \leq \|f(a+h) - f(a) - L_1 h\| + \|f(a+h) - f(a) - L_2 h\| = o(\|h\|)$. For any fixed $v$, put $h = tv$ and let $t \to 0$: $\|(L_1 - L_2)v\| \leq o(|t|)/|t| \to 0$, so $(L_1 - L_2)v = 0$ for all $v$, and $L_1 = L_2$. For the entries, put $h = t e_j$ in the definition applied to the $i$-th component:

$$\frac{f_i(a + t e_j) - f_i(a)}{t} - (L e_j)_i = \frac{o(|t|)}{|t|} \to 0,$$

so $\partial f_i / \partial x_j (a) = \lim_{t \to 0} (f_i(a + t e_j) - f_i(a))/t = (L e_j)_i$. $\blacksquare$

Caution (partials do not a derivative make). The converse of Proposition 1.1 fails. Let $f(x, y) = x^2 y / (x^4 + y^2)$ for $(x, y) \neq (0, 0)$ and $f(0, 0) = 0$. Both partials at the origin vanish (on either axis the function is identically $0$). Yet $f$ is not differentiable at the origin: the derivative would have to be the zero map (Proposition 1.1), requiring $f(h)/\|h\| \to 0$; but along $h = (t, t^2)$ we get $f(h) = t^4/(t^4 + t^4) = 1/2$ while $\|h\| \to 0$, so $f(h)/\|h\| \to \infty$. Stronger still, every directional derivative at the origin exists and is $0$ (for fixed $h$, $f(th) = t\, h_1^2 h_2 / (t^2 h_1^4 + h_2^2) \to 0$ when $h_2 \neq 0$, and is $0$ when $h_2 = 0$), and the function is still not differentiable. $\blacksquare$

Proposition 1.2 (algebra of derivatives). Sums and scalar multiples of differentiable maps are differentiable, with the obvious derivatives. If $g, h: \mathbb{R}^n \to \mathbb{R}$ are differentiable at $a$, the product $gh$ is differentiable at $a$ with derivative $h(a) g'(a) + g(a) h'(a)$. If $g$ is differentiable at $a$ with $g(a) \neq 0$, the reciprocal $1/g$ is differentiable at $a$ with derivative $-g'(a)/g(a)^2$, and $1/g$ is continuous at $a$.

Proof. Sums and scalars are immediate from linearity of the definition. Product: expand

$$(g h)(a{+}h) - (g h)(a) - \big(h(a) g'(a) + g(a) h'(a)\big) h = \Delta_g \, h(a) + g(a)\, \Delta_h + \Delta_g \Delta_h,$$

with $\Delta_g = g(a{+}h) - g(a) - g'(a)h = o(\|h\|)$ and $\Delta_h$ likewise. Dividing by $\|h\|$: the first term dies because $\Delta_g = o(\|h\|)$ and $h(a)$ is a constant; the third dies as $o(\|h\|) \cdot o(\|h\|)$; and the middle dies since $g(a)$ is a constant. (For vector-valued factors, apply componentwise.)

Reciprocal: with everything evaluated at $a$ and $a{+}h$, a direct computation gives

$$\frac{1}{g(a{+}h)} - \frac{1}{g(a)} + \frac{g'(a)\,h}{g(a)^2} = \frac{-\Delta_g \; g(a) + g'(a) h\, \big(g(a{+}h) - g(a)\big)} {g(a{+}h)\, g(a)^2}.$$

Since $g$ is continuous at $a$ (differentiability implies continuity, the definition with $Lh$ subtracted), $g(a{+}h) \to g(a) \neq 0$, so the denominator stays bounded away from $0$ near $h = 0$; the numerator is $o(\|h\|)$ plus (bounded linear in $h$) times $o(1)$: divided by $\|h\|$ it tends to $0$. Continuity of $1/g$ at $a$ is the same computation with the derivative term dropped. $\blacksquare$

Corollary 1.3 (polynomials and rational functions are smooth). Every polynomial map $\mathbb{R}^n \to \mathbb{R}$ is smooth; every rational function is smooth on the open set where its denominator is nonzero.

Proof. Each coordinate function $u_i$ is linear, hence differentiable with derivative itself (the definition’s error term is identically $0$). A monomial is a finite product of coordinate functions, smooth by repeated application of the product rule (Proposition 1.2); polynomials are sums of monomials. A rational function is a quotient of polynomials; on its domain the denominator never vanishes, and the reciprocal rule applies. $\blacksquare$

Lemma 1.4 (pairing lemma). If $g: \mathbb{R}^n \to \mathbb{R}^k$ and $h: \mathbb{R}^m \to \mathbb{R}^{\ell}$ are differentiable at the relevant points, then $(g, h)(x, y) = (g(x), h(y))$ is differentiable at $(a, b)$ with derivative the block map $(u, v) \mapsto (g'(a)u,\, h'(b)v)$.

Proof. $\|(g,h)(a{+}u, b{+}v) - (g,h)(a,b) - (g'(a)u, h'(b)v)\|^2 = \|g(a{+}u) - g(a) - g'(a)u\|^2 + \|h(b{+}v) - h(b) - h'(b)v\|^2$ (squares separate the coordinates), and each term is $o(\|(u,v)\|^2)$ by the differentiability of $g$ and $h$. $\blacksquare$

Lemma 1.5 (smoothness is local). If a map is differentiable at every point of an open set $U$, then it is differentiable at every point of any open subset of $U$. Conversely, if each point of an open set has some neighbourhood on which the map is differentiable at every point, then the map is differentiable at every point of the set.

Proof. Differentiability at $a$ is a statement about the values of $f$ on arbitrarily small neighbourhoods of $a$: restricting the domain to a smaller open set changes nothing at interior points. Both directions are this one remark. $\blacksquare$

Quick check.

1. Define the derivative of $f$ at $a$.
2. Do partial derivatives at a point determine the derivative?
3. Does existence of all partial derivatives force differentiability?
4. What is the derivative of the product $gh$ at $a$?
5. Which functions were proved smooth with no calculus input?

Answers: 1. The unique linear $L$ with $\|f(a+h) - f(a) - Lh\| = o(\|h\|)$. 2. Yes, when the derivative exists: the entries are the partials (Proposition 1.1). 3. No: the Caution’s counterexample has every directional derivative and no derivative. 4. $h(a)\, g'(a) + g(a)\, h'(a)$ (Proposition 1.2). 5. Polynomials and rational functions on their domains (Corollary 1.3).

Definition. A base for a topology on $X$ is a collection $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. The space is second countable if it has a countable base.

Proposition 2.1. $\mathbb{R}^n$ is second countable; every subspace of a second countable space is second countable; every subspace of a Hausdorff space is Hausdorff.

Proof. $\mathbb{R}^n$: the balls $B(q, r)$ with $q \in \mathbb{Q}^n$ and $r \in \mathbb{Q}$, $r > 0$, form a countable collection ($\mathbb{Q}^n$ is countable by finite products of countables). They base the topology: given an open $U$ and $x \in U$, some ball $B(x, \varepsilon) \subseteq U$; by density of $\mathbb{Q}$ pick $q \in \mathbb{Q}^n$ with $\|x - q\| < \varepsilon/3$ and a rational $r$ with $\|x - q\| < r < 2\varepsilon/3$; then $x \in B(q, r)$ and $B(q, r) \subseteq B(x, \varepsilon)$ (for $z \in B(q, r)$: $\|z - x\| \leq \|z - q\| + \|q - x\| < r + \varepsilon/3 < \varepsilon$), so every point of $U$ is covered by members inside $U$: $U$ is a union of such balls. Subspaces: if $\mathcal{B}$ is a countable base of $X$ and $S \subseteq X$, then $\{B \cap S : B \in \mathcal{B}\}$ is a countable base of $S$ (an open set of $S$ is $S \cap W'$ with $W' = \bigcup B_i$, and $S \cap W' = \bigcup (B_i \cap S)$). Hausdorff subspaces: for $x \neq y$ in $S$, the separating opens $U \ni x$, $V \ni y$ in $X$ intersect $S$ in disjoint opens of the subspace. $\blacksquare$

Definition. A topological manifold of dimension $n$ is a Hausdorff, second countable space in which every point has an open neighbourhood homeomorphic to an open subset of $\mathbb{R}^n$.

Example (the model case). $\mathbb{R}^n$ itself: Hausdorff and second countable (Propositions 5.1 and 2.1 of Level 2 and Proposition 2.1 above), and the identity map is a chart. $\blacksquare$

Example (a non-manifold: the line with two origins). Let $M = (\mathbb{R} \setminus \{0\}) \cup \{p, q\}$, where $p \neq q$ are two extra points, and declare the following collection $\mathcal{B}$ of subsets to be a base: all sets $N(t, \varepsilon) = (t - \varepsilon, t + \varepsilon) \setminus \{0\}$ for $t \neq 0$; the sets $P(\varepsilon) = \{p\} \cup \big((-\varepsilon, 0) \cup (0, \varepsilon)\big)$; and the sets $Q(\varepsilon)$ defined likewise with $q$. (Base check: pairwise intersections of members are unions of members – intervals meet in intervals, $P(\varepsilon) \cap P(\delta) = P(\min)$, and a punctured interval meets a punctured interval in punctured intervals, each point of which carries a small ball avoiding $0$.) With the topology $\mathcal{B}$ generates, every point of $M$ has a Euclidean neighbourhood: $t \neq 0$ has $N(t, \varepsilon) \cong (t - \varepsilon, t + \varepsilon)$, and $p$ has $\varphi: P(\varepsilon) \to (-\varepsilon, \varepsilon)$, $\varphi(p) = 0$, $\varphi(s) = s$ – a homeomorphism (the preimage of any interval is a member of $\mathcal{B}$ or a union of them), and likewise for $q$. But $M$ is not Hausdorff: $P(\varepsilon) \ni p$ and $Q(\delta) \ni q$ always intersect, in $(-\min(\varepsilon, \delta), 0) \cup (0, \min(\varepsilon, \delta))$. So the Hausdorff clause in the definition is not decoration. $\blacksquare$

Proposition 2.2 (open submanifolds). An open subset of an $n$-dimensional topological manifold is an $n$-dimensional topological manifold.

Proof. Hausdorff and second countable pass to subspaces (Proposition 2.1). If $\varphi: U \to \varphi(U)$ is a chart and $W \subseteq U$ is open in the manifold, then $W$ is open in $U$ (the subspace topologies of the open $U$ agree), so $\varphi(W)$ is open in $\varphi(U)$, hence open in $\mathbb{R}^n$, and $\varphi|_W: W \to \varphi(W)$ is a homeomorphism: charts cover $W$. $\blacksquare$

Quick check.

1. Name the three clauses defining a topological manifold.
2. What is a base, and what does second countable mean?
3. Why is the line with two origins not a manifold?
4. Is an open subset of a manifold a manifold?
5. Why is $\mathbb{R}^n$ second countable?

Answers: 1. Hausdorff; second countable; locally Euclidean of dimension $n$. 2. A base is opens whose unions give all opens; second countable means a countable base. 3. It is not Hausdorff: the two origins cannot be separated. 4. Yes, with the restricted charts (Proposition 2.2). 5. Balls with rational centres and rational radii (Proposition 2.1).

Definition. A chart of a topological manifold $M$ is a homeomorphism $\varphi: U \to \varphi(U)$ from an open $U \subseteq M$ onto an open subset of $\mathbb{R}^n$. An atlas is a set of charts whose domains cover $M$. For two charts $(U, \varphi)$ and $(V, \psi)$, the transition map is

$$\psi \circ \varphi^{-1}: \varphi(U \cap V) \to \psi(U \cap V),$$

defined on an open subset of $\mathbb{R}^n$ (the image of the open $U \cap V$ under the homeomorphism $\varphi$). The charts are smoothly compatible if both transition directions are smooth (all derivatives existing at every point of their open domains, Section 1). An atlas is a smooth atlas if its charts are pairwise smoothly compatible.

Definition. A smooth atlas $\mathcal{A}$ is maximal if no strictly larger atlas containing it is smooth. A smooth structure on $M$ is a maximal smooth atlas, and $M$ together with one is a smooth manifold.

Nothing yet guarantees maximal atlases exist or are unique; the next two results supply both. The heart is that smooth compatibility is not transitive for arbitrary triples of charts (a counterexample follows the proof), but transitivity does hold through a common atlas, locally.

Proposition 3.1 (transitivity through a common atlas). Let $\mathcal{A}$ be a smooth atlas and let $(U, \varphi)$, $(V, \psi)$ be two charts, each smoothly compatible with every chart of $\mathcal{A}$. Then $(U, \varphi)$ and $(V, \psi)$ are smoothly compatible with each other.

Proof. Let $y \in \varphi(U \cap V)$, and let $p = \varphi^{-1}(y) \in U \cap V$. Choose a chart $(W, \theta) \in \mathcal{A}$ with $p \in W$ (the domains of $\mathcal{A}$ cover $M$). The set $\varphi(U \cap V \cap W)$ is an open neighbourhood of $y$ (image of an open set under the homeomorphism $\varphi$), and on it,

$$\psi \circ \varphi^{-1} = (\psi \circ \theta^{-1}) \circ (\theta \circ \varphi^{-1}),$$

where the right side is defined: $\theta \circ \varphi^{-1}$ carries $\varphi(U \cap V \cap W)$ into $\theta(U \cap V \cap W) \subseteq \theta(W \cap V)$, whose image under $\psi \circ \theta^{-1}$ lies in $\psi(U \cap V)$; both factors are smooth (each chart is compatible with every chart of $\mathcal{A}$), and by the chain rule the composition is smooth at $y$. The same argument with the roles of $\varphi$ and $\psi$ exchanged handles the inverse transition. Since $y \in \varphi(U \cap V)$ was arbitrary, both transition maps are differentiable at every point of their domains (Lemma 1.5): the charts are smoothly compatible. $\blacksquare$

Why raw transitivity fails. On the topological manifold $\mathbb{R}$, let $\psi$ be the identity chart on $(0, 1)$, $\varphi$ the identity chart on $(0, 3)$, and let $\theta: (0, 3) \to (1/4,\, 4)$ be $\theta(t) = t + \frac{1}{4}|t - 1|^3$. Then $\theta$ is a homeomorphism onto its (open) image: it is continuous and strictly increasing, since its derivative is $1 - \frac{3}{4}(t - 1)^2 \geq \frac{1}{4} > 0$, and a continuous strictly monotone map on an interval has a continuous inverse (the inverse is monotone on an interval, and the one-line $\varepsilon$–$\delta$ bookkeeping is standard). It is smoothly compatible with $\psi$: on the overlap $(0, 1)$ the transition $t \mapsto \theta^{-1}(t)$ is smooth, because $\theta$ is smooth there (a polynomial in $t$) with derivative never zero, so the inverse function theorem gives a smooth inverse. And $\varphi$ is compatible with $\psi$ without any computation (both are restrictions of the identity). But $\varphi$ and $\theta$ are not compatible: the transition $\theta \circ \varphi^{-1} = \theta$ is twice but not three times differentiable at $t = 1$ ($|t-1|^3$ has a third derivative that jumps there). The failure in Proposition 3.1’s proof is exactly the point $p = 1$: no chart of the atlas $\{\psi\}$ contains it. Transitivity through a covering atlas repairs this. $\blacksquare$

Theorem 3.2 (existence and uniqueness of the maximal atlas). Every smooth atlas $\mathcal{A}$ on $M$ is contained in exactly one maximal smooth atlas, namely

$$\operatorname{Max}(\mathcal{A}) = \{\,\text{charts smoothly compatible with every chart of } \mathcal{A}\,\}.$$

Proof. It is an atlas: it contains $\mathcal{A}$ (each chart of $\mathcal{A}$ is compatible with every chart of $\mathcal{A}$), so its domains cover $M$. It is smooth: any two of its members are compatible with each other by Proposition 3.1 (with the common atlas $\mathcal{A}$). It is maximal: if a chart $(U, \varphi)$ is smoothly compatible with every chart of $\operatorname{Max}(\mathcal{A})$, then in particular with every chart of $\mathcal{A} \subseteq \operatorname{Max}(\mathcal{A})$, so it already lies in $\operatorname{Max}(\mathcal{A})$ by definition; hence no chart can be adjoined. It is unique: let $\mathcal{A}_1$ be any maximal smooth atlas containing $\mathcal{A}$. Every chart of $\mathcal{A}_1$ is compatible with every chart of $\mathcal{A}$ (pairwise compatibility inside $\mathcal{A}_1$, which contains $\mathcal{A}$), so $\mathcal{A}_1 \subseteq \operatorname{Max}(\mathcal{A})$ by definition of the latter. Conversely, take $\varphi \in \operatorname{Max}(\mathcal{A})$ and any $\theta \in \mathcal{A}_1$: both are compatible with every chart of $\mathcal{A}$, so Proposition 3.1 (common atlas $\mathcal{A}$) makes them compatible with each other. Hence $\mathcal{A}_1 \cup \{\varphi\}$ is a smooth atlas, and maximality of $\mathcal{A}_1$ forces $\varphi \in \mathcal{A}_1$. So $\operatorname{Max}(\mathcal{A}) \subseteq \mathcal{A}_1$, and the two are equal. $\blacksquare$

The One-Line Reveal. The maximal atlas is defined into existence: “compatible with everything in sight” is self-closing, and Proposition 3.1 is what makes the resulting family pairwise consistent. Uniqueness is the same proposition run in the other direction.

Definition. A smooth manifold is a topological manifold with a smooth structure; charts of the maximal atlas are admissible. A subset is open in the manifold exactly when it is open in the underlying topology: the smooth structure adds no topology.

Quick check.

1. What is a chart, and what is an atlas?
2. What is the transition map of charts $(U, \varphi)$, $(V, \psi)$?
3. What does smoothly compatible mean?
4. Is smooth compatibility transitive among arbitrary charts?
5. What repairs transitivity, and how does Theorem 3.2 use it?

Answers: 1. A homeomorphism onto an open subset of $\mathbb{R}^n$; an atlas is charts covering $M$. 2. $\psi \circ \varphi^{-1}$ on $\varphi(U \cap V)$. 3. Both transition directions are smooth. 4. No: the $\theta(t) = t + \frac{1}{4}|t-1|^3$ counterexample is compatible with $\psi$ but not with $\varphi$. 5. Compatibility through a common covering atlas, applied locally (Proposition 3.1); that is what makes $\operatorname{Max}(\mathcal{A})$ consistent and unique (Theorem 3.2).

Example (the sphere, entirely by algebra). Let

$$S^n = \{(u_1, \dots, u_{n+1}) \in \mathbb{R}^{n+1} : u_1^2 + \cdots + u_{n+1}^2 = 1\}$$

with its subspace topology, where $\mathbb{R}^{n+1}$ carries the product topology of Level 2, Section 4. Let $N = (0, \dots, 0, 1)$ and $S = (0, \dots, 0, -1)$ be the poles. Define the stereographic projections

$$\varphi_N: S^n \setminus \{N\} \to \mathbb{R}^n, \qquad \varphi_N(u) = \frac{(u_1, \dots, u_n)}{1 - u_{n+1}},$$ $$\varphi_S: S^n \setminus \{S\} \to \mathbb{R}^n, \qquad \varphi_S(u) = \frac{(u_1, \dots, u_n)}{1 + u_{n+1}}.$$

They are well defined: $1 - u_{n+1} = 0$ on the sphere only at $N$ ($u_{n+1} = 1$ forces all other coordinates to $0$), and likewise $1 + u_{n+1} = 0$ only at $S$. They are mutual inverses as maps to and from the sphere: the inverse of $\varphi_N$ is

$$\psi_N(y) = \left(\frac{2y}{\|y\|^2 + 1},\; \frac{\|y\|^2 - 1}{\|y\|^2 + 1}\right),$$

and the verification is pure algebra. For $u \in S^n \setminus \{N\}$ put $y = \varphi_N(u) = u'/(1 - u_{n+1})$ with $u' = (u_1, \dots, u_n)$. On the sphere $\|u'\|^2 = 1 - u_{n+1}^2 = (1 - u_{n+1})(1 + u_{n+1})$, so

$$\|y\|^2 + 1 = \frac{(1 - u_{n+1})(1 + u_{n+1}) + (1 - u_{n+1})^2}{(1 - u_{n+1})^2} = \frac{2}{1 - u_{n+1}},$$

and therefore $\frac{2y}{\|y\|^2 + 1} = u'$ and $\frac{\|y\|^2 - 1}{\|y\|^2 + 1} = \frac{(1 - u_{n+1})(1 + u_{n+1}) - (1 - u_{n+1})^2}{2(1 - u_{n+1})} = u_{n+1}$: the composite recovers $u$. The composite $\varphi_N \circ \psi_N$ is the identity by the same identity read backwards ($1 - u_{n+1} = 2/(\|y\|^2 + 1)$ for $u = \psi_N(y)$). The same formulas with poles swapped handle $\varphi_S$. They are homeomorphisms: both directions are continuous (Corollary 1.3: rational functions, with denominators bounded away from $0$), and continuous mutual inverses are homeomorphisms. $S^n$ is a topological manifold: it is Hausdorff and second countable as a subspace of $\mathbb{R}^{n+1}$ (Proposition 2.1), and the two chart domains $S^n \setminus \{N\}$, $S^n \setminus \{S\}$ cover it. It is a smooth manifold: the transition map

$$\varphi_S \circ \psi_N(y) = \frac{2y/(\|y\|^2 + 1)}{1 + (\|y\|^2 - 1)/(\|y\|^2 + 1)} = \frac{y}{\|y\|^2},$$

(using $1 + u_{n+1} = 2\|y\|^2/(\|y\|^2 + 1)$ for $u = \psi_N(y)$) is smooth on $\mathbb{R}^n \setminus \{0\}$, which is exactly $\varphi_N(U_N \cap U_S)$, and it is its own inverse, so the inverse transition is the same smooth map. The atlas $\{(\,S^n \setminus \{N\}, \varphi_N),\, (S^n \setminus \{S\}, \varphi_S)\,\}$ is smooth, and Theorem 3.2 gives the maximal one. Note what did not happen: no trigonometry, no limits beyond Section 1’s algebra. $\blacksquare$

Theorem 4.1 (products). If $M, N$ are smooth manifolds of dimensions $m, n$, then $M \times N$ with the product topology is a smooth manifold of dimension $m + n$, with product charts $(U \times V, \varphi \times \psi)$.

Proof. Topology. $M \times N$ is Hausdorff: for $(x, y) \neq (x', y')$, one coordinate differs, say $x \neq x'$, and disjoint neighbourhoods $U \ni x$, $U' \ni x'$ separate the points as boxes $U \times N$, $U' \times N$. Second countable: if $\mathcal{B}_M, \mathcal{B}_N$ are countable bases, the boxes $\{B \times C\}$ over them are countable and base the product (Level 2, Section 4: product opens are unions of boxes, and each box’s factors are unions of base members). Charts. For charts $(U, \varphi)$ of $M$ and $(V, \psi)$ of $N$, the product map $\varphi \times \psi: U \times V \to \varphi(U) \times \psi(V)$ is a homeomorphism between open sets (product of homeomorphisms, with target open in $\mathbb{R}^{m+n} = \mathbb{R}^m \times \mathbb{R}^n$), and such charts cover $M \times N$. Compatibility. The transition between $(U_\alpha \times V_\alpha, \varphi_\alpha \times \psi_\alpha)$ and $(U_\beta \times V_\beta, \varphi_\beta \times \psi_\beta)$ is

$$(\varphi_\beta \circ \varphi_\alpha^{-1}) \;\times\; (\psi_\beta \circ \psi_\alpha^{-1}),$$

since composing and inverting act coordinatewise; it is smooth by the pairing lemma (Lemma 1.4) applied to the two smooth transition maps. So the product charts form a smooth atlas, and Theorem 3.2 extends it. $\blacksquare$

Quick check.

1. Write the stereographic inverse $\psi_N$.
2. What is the transition $\varphi_S \circ \psi_N$?
3. Why is $S^n$ Hausdorff and second countable?
4. Why is $M \times N$ a manifold, and of what dimension?
5. What did the sphere construction not need?

Answers: 1. $\psi_N(y) = (2y/(\|y\|^2+1),\ (\|y\|^2-1)/(\|y\|^2+1))$. 2. $y \mapsto y/\|y\|^2$ on $\mathbb{R}^n \setminus \{0\}$. 3. It is a subspace of $\mathbb{R}^{n+1}$ (Proposition 2.1). 4. Product charts form a smooth atlas (Theorem 4.1); dimension $m + n$. 5. No trigonometry: only rational functions and algebra.

Definition. A map $f: M \to N$ of smooth manifolds is smooth at $p \in M$ if there exist an admissible chart $(U, \varphi)$ of $M$ at $p$ and an admissible chart $(V, \psi)$ of $N$ at $f(p)$ such that

1. $f(U) \subseteq V$, and
2. the chart representation $\psi \circ f \circ \varphi^{-1}: \varphi(U) \to \psi(V)$ is smooth on the open set $\varphi(U)$.

The map is smooth if it is smooth at every point. Note the asymmetry with the topological definition of continuity: smoothness demands a chart around $p$ whose entire image under $f$ sits inside the target chart. The demand is satisfiable precisely because $U$ may be shrunk, and the next two propositions show the notion is well defined and implies continuity.

Proposition 5.1 (smoothness implies continuity). A map that is smooth at $p$ is continuous at $p$.

Proof. With $U, \varphi, V, \psi$ as in the definition, $f$ factors near $p$ as $f = \psi^{-1} \circ (\psi \circ f \circ \varphi^{-1}) \circ \varphi$ on $U$: for $x \in U$, $f(x) = \psi^{-1}\big((\psi \circ f \circ \varphi^{-1})(\varphi(x))\big)$, since $\varphi(x) \in \varphi(U)$ and the representation takes values in $\psi(V)$. The right side is a composition of continuous maps ($\varphi$ and $\psi^{-1}$ are homeomorphisms; the middle is smooth on an open set, hence continuous), so $f$ is continuous at $p$. $\blacksquare$

Proposition 5.2 (well-definedness). If $f$ is smooth at $p$ via one pair of charts, then for every admissible chart $(U_2, \varphi_2)$ at $p$ and $(V_2, \psi_2)$ at $f(p)$ there is an open $U' \subseteq U_2$ containing $p$ with $f(U') \subseteq V_2$ and $\psi_2 \circ f \circ \varphi_2^{-1}$ smooth on $\varphi_2(U')$: the definition does not depend on the charts used.

Proof. Let $(U_1, \varphi_1)$, $(V_1, \psi_1)$ witness smoothness. By Proposition 5.1, $f$ is continuous at $p$, so $U' := U_1 \cap U_2 \cap f^{-1}(V_2)$ is an open neighbourhood of $p$ (a preimage of the open $V_2$ under a map continuous at $p$, intersected with two open sets). Restricting $U_1$ to $U'$ leaves a legitimate chart (a homeomorphism onto an open subset) and satisfies the definition’s demands. Now on $\varphi_2(U')$,

$$\psi_2 \circ f \circ \varphi_2^{-1} = (\psi_2 \circ \psi_1^{-1}) \circ (\psi_1 \circ f \circ \varphi_1^{-1}) \circ (\varphi_1 \circ \varphi_2^{-1}),$$

and the composite is defined at every point of $\varphi_2(U')$: for $x \in U'$ we have $f(x) \in V_1$ (since $x \in U_1$ and $f(U_1) \subseteq V_1$) and $f(x) \in V_2$, so the middle factor is applied at $\varphi_1(x)$, the outer factor at $\psi_1(f(x)) \in \psi_1(V_1 \cap V_2)$ (exactly the domain of the transition $\psi_2 \circ \psi_1^{-1}$), and the inner factor at $\varphi_2(x) \in \varphi_2(U_1 \cap U_2)$. Each factor is smooth on its open domain (transitions by admissibility, the middle by hypothesis), so the chain rule makes the composition smooth on the open $\varphi_2(U')$. $\blacksquare$

Proposition 5.3 (closure properties). Smooth maps are continuous. Compositions of smooth maps are smooth. Identity maps are smooth. Affine maps are smooth, with derivative the linear part.

Proof. Continuity is Proposition 5.1 at each point. Composition: let $f: M \to N$ be smooth at $p$ and $g: N \to P$ smooth at $f(p)$, witnessed by charts $(U_1, \varphi_1)$, $(V_1, \psi_1)$ for $f$ and $(U_2, \varphi_2)$, $(V_2, \psi_2)$ for $g$ (so $f(U_1) \subseteq V_1$ and $g(U_2) \subseteq V_2$). By Proposition 5.2 applied to $f$ with the target chart $(U_2, \varphi_2)$ of $N$, there is an open $U' \subseteq U_1 \cap U_2$ containing $p$ with $f(U') \subseteq U_2$ and $\varphi_N$-representation $\varphi_2 \circ f \circ \varphi_1^{-1}$ smooth on $\varphi_1(U')$. Then the representation of $g \circ f$ through $(U_1, \varphi_1)$ and $(V_2, \psi_2)$ is

$$\psi_2 \circ g \circ f \circ \varphi_1^{-1} = (\psi_2 \circ g \circ \varphi_2^{-1}) \circ (\varphi_2 \circ f \circ \varphi_1^{-1}),$$

defined on $\varphi_1(U')$ (for $x \in U'$: $f(x) \in U_2$, so the inner factor applies and lands in $\varphi_2(U_2)$, where the outer factor is smooth and takes values in $\psi_2(V_2)$), and smooth by the chain rule. Identity: the representation through any chart $(U, \varphi)$ is $\varphi \circ \operatorname{id} \circ \varphi^{-1}$, the identity of the open set $\varphi(U)$: linear, so smooth by the affine case below. Affine maps: for $F(x) = Lx + b$, the error in the derivative definition is $\|F(a + h) - F(a) - Lh\| = 0$ identically, so $F'(a) = L$ at every point. $\blacksquare$

Definition. A diffeomorphism $f: M \to N$ is a bijection that is smooth with smooth inverse; the manifolds are then diffeomorphic, which is the smooth analogue of homeomorphism.

Theorem 5.4 (dimension is a diffeomorphism invariant). If diffeomorphic smooth manifolds have dimensions $n$ and $m$, then $n = m$.

Proof. Let $f: M \to N$ be a diffeomorphism, $p \in M$, and pick charts $(U, \varphi)$ at $p$ and $(V, \psi)$ at $f(p)$, shrinking $U$ so that $f(U) \subseteq V$ (possible by Proposition 5.1’s continuity). The chart representation $F = \psi \circ f \circ \varphi^{-1}$ is then a diffeomorphism between the open subsets $\varphi(U) \subseteq \mathbb{R}^n$ and $\psi(f(U)) \subseteq \mathbb{R}^m$: it is smooth by hypothesis, bijective (with inverse $\varphi \circ f^{-1} \circ \psi^{-1}$, which is smooth by the same transition-composition argument as Proposition 5.2 applied to $f^{-1}$, whose chart representations are smooth by the definition of a diffeomorphism). Fix $x \in \varphi(U)$. The chain rule applied to $F^{-1} \circ F = \operatorname{id}$ and $F \circ F^{-1} = \operatorname{id}$ gives, for $A = F'(x)$ ($m \times n$) and $B = (F^{-1})'(F(x))$ ($n \times m$): $BA = I_n$ and $AB = I_m$. Then $\operatorname{im}(AB) = \mathbb{R}^m \subseteq \operatorname{im} A$, so $A$ is surjective, and by rank–nullity (Level 1B, Theorem 1.7) $m \leq n$; symmetrically $BA = I_n$ makes $B$ surjective and gives $n \leq m$. Hence $n = m$. $\blacksquare$

This is why the definition of a manifold fixes one dimension $n$ per component and never needs to defend it: the inverse function theorem plus rank–nullity make the dimension unforgeable.

Quick check.

1. When is $f: M \to N$ smooth at $p$?
2. How many lines does smoothness-to-continuity take, and how?
3. Why is the definition independent of charts?
4. What is a diffeomorphism?
5. Why is dimension a diffeomorphism invariant?

Answers: 1. Some chart at $p$ carries inside a chart at $f(p)$ with a smooth chart representation. 2. Two lines: factor $f$ as $\psi^{-1}$ after representation after $\varphi$ on the chart domain (Proposition 5.1). 3. Continuity lets domains shrink; transitions are smooth; the chain rule finishes (Proposition 5.2). 4. A smooth bijection with smooth inverse. 5. The chain rule gives $BA = I_n$ and $AB = I_m$; rank–nullity forces $n = m$ (Theorem 5.4).

1. Prove that the line with two origins is not Hausdorff, and conclude which clause of the manifold definition rules it out. (§2 – any two neighbourhoods of the two origins contain a common punctured interval.)
2. Verify $\psi_N(\varphi_N(u)) = u$ for the stereographic chart, doing every algebraic step. (§4 – the key identity is $\|u'\|^2 = (1 - u_{n+1})(1 + u_{n+1})$ on the sphere.)
3. Prove that the open interval $(a, b)$ is a smooth manifold of dimension
   1. (§2, §4 – subspace of Hausdorff second countable $\mathbb{R}$; identity chart.)
4. Prove that every directional derivative of the Caution function at the origin exists and equals $0$, and that the function is nevertheless not differentiable there. (§1 – the counterexample in full.)
5. Prove the special case of Proposition 3.1 in which the point $p$ lies in all three chart domains. (§3 – the composition of the two transition maps, by the chain rule.)
6. Show that the atlas $\{\operatorname{id}_{\mathbb{R}}\}$ and the atlas $\{t \mapsto t^3\}$ on $\mathbb{R}$ generate different maximal atlases. (§3 – the transition between the two charts is $t \mapsto t^3$ with inverse $t \mapsto t^{1/3}$, not differentiable at $0$; Proposition 3.1 does not force compatibility because… check: does it? Apply the proposition and see why it does not apply.)
7. Prove that a diffeomorphism $f: M \to N$ restricts, for every open $U \subseteq M$, to a diffeomorphism $U \to f(U)$. (§5, §2 – $f(U)$ is open by continuity, and charts restrict as in Proposition 2.2.)

Problems. The quick checks above tested recall. These chain several results and require full written arguments; hints are included where the jump is large.

8. Prove that every chart is a diffeomorphism onto its open image, viewed as an open submanifold of $\mathbb{R}^n$. (Its chart representation is the identity; Propositions 2.2 and 5.3 finish.)
9. Prove that $\gamma: (0, 2\pi) \to S^1 \setminus \{(1, 0)\}$, $\gamma(t) = (\cos t, \sin t)$, is a diffeomorphism onto its image. (Bijective by trigonometry; smooth by Proposition 4.2; the inverse has nonzero derivative, so the inverse function theorem smooths it.)

The entire page follows from one decision: a manifold’s smoothness is not a property of the space but a choice of coordinate bookkeeping, made coherent by requiring all coordinate changes to be smooth. The coherence proof is local (Proposition 3.1 plus Lemma 1.5), which is precisely why the naive transitivity claim fails and why the maximal atlas still works: consistency only ever needs to be checked through charts you already trust.

• Level 3B: Tangent Spaces – tangent vectors as derivations on germs, the differential, and the inverse function theorem on manifolds, built directly on the chart machinery of this page.
• Level 2: Topology – the Hausdorff, second-countable, subspace and product theorems used throughout this page, with proofs.
• An Introduction to Manifolds – Loring Tu’s book, whose Chapters 1–2 develop this page’s material with the worked details of multivariable calculus assumed here as input.