Smooth Manifolds and Smooth Maps: Examples and Counterexamples

Taking $U={\mathbb{R}}^n$, the identity chart exhibits ${\mathbb{R}}^n$ as a topological $n$-manifold and [F1] supplies its smooth structure. Hence ${\mathbb{R}}^n$ is a smooth $n$-manifold by [F2].

For a general open subset $U\subseteq {\mathbb{R}}^n$, [F1] states exactly that $U$ inherits the standard smooth structure, so again [F2] makes $U$ a smooth $n$-manifold.

The domains $S^1\setminus \{N\}$ and $S^1\setminus \{S\}$ are open in $S^1$ [given] and cover it. The displayed inverse formulas show that each ${\sigma }_N$ and ${\sigma }_S$ is a homeomorphism onto $\mathbb{R}$: composing the inverse with the chart returns the original point, and composing the chart with the inverse gives $u$. So these are genuine charts.

On the overlap $S^1\setminus \{N,S\}$ the transition maps are

This smooth atlas equips $S^1$ with a smooth structure, so [F3] makes the [F3, step 2.1] circle a smooth $1$-manifold.

The domains $S^n\setminus \{N\}$ and $S^n\setminus \{S\}$ are open and cover [given] $S^n$. The inverse formulas ${\sigma }_N^{-1}(u)=(\frac{2u}{1+\mathrm{\parallel }u{\mathrm{\parallel }}^2},\frac{\mathrm{\parallel }u{\mathrm{\parallel }}^2-1}{1+\mathrm{\parallel }u{\mathrm{\parallel }}^2}),{\sigma }_S^{-1}(u)=(\frac{2u}{1+\mathrm{\parallel }u{\mathrm{\parallel }}^2},\frac{1-\mathrm{\parallel }u{\mathrm{\parallel }}^2}{1+\mathrm{\parallel }u{\mathrm{\parallel }}^2})$ show that both maps are homeomorphisms onto ${\mathbb{R}}^n$.

On the overlap the transition maps are ${\sigma }_S\circ {\sigma }_N^{-1}(u)=\frac{u}{\mathrm{\parallel }u{\mathrm{\parallel }}^2},{\sigma }_N\circ {\sigma }_S^{-1}(u)=\frac{u}{\mathrm{\parallel }u{\mathrm{\parallel }}^2},$ defined on ${\mathbb{R}}^n\setminus \{0\}$, so they are smooth rational maps there. Hence the two stereographic charts are smoothly compatible by [F2], and [F1] makes them a smooth atlas on $S^n$.

Therefore $S^n$ is a smooth manifold by [F3].

The sets $U_i$ cover ${\mathbb{R}\mathbb{P}}^n$ because every nonzero vector in [F1] ${\mathbb{R}}^{n+1}$ has at least one nonzero coordinate. Each $U_i$ is open by [F1], since its preimage is $\{x_i\mathrm{\ne }0\}\subseteq {\mathbb{R}}^{n+1}\setminus \{0\}$. The inverse chart sends $(u_1,\dots ,u_n)\in {\mathbb{R}}^n$ to the projective class with $i$-th coordinate $1$ and the remaining coordinates given by the $u_j$, so each ${\varphi }_i$ is a homeomorphism $U_i\to {\mathbb{R}}^n$.

On $U_i\cap U_j$ the transition map ${\varphi }_j\circ {\varphi }_i^{-1}$ is obtained by [F2, step 1.1] dividing all affine coordinates by the coordinate corresponding to $x_j\mathrm{/}{x}_i$, which is nonzero on the overlap. Thus every transition function is rational with nonvanishing denominator on its domain, hence smooth. Therefore the family $(U_i,{\varphi }_i)$ is a smooth atlas by [F2].

This atlas equips ${\mathbb{R}\mathbb{P}}^n$ with a smooth structure, so [F3] makes [F3, step 2.1] ${\mathbb{R}\mathbb{P}}^n$ a smooth $n$-manifold.

By [F1], each factor $S^1$ is a smooth $1$-manifold.

Applying [F2] to the two circle factors gives a canonical product smooth [F2, step 1.1] structure on $S^1\times S^1$, making it a smooth $(1+1)$-manifold. This is the torus $T^2$.

By [F1], every summand ${\mathbb{R}}_m$ is a smooth $1$-manifold.

The index set $\mathbb{N}$ is countable, so [F2] applies to the family [F2, step 1.1] $({\mathbb{R}}_m{)}_{m\in \mathbb{N}}$ and yields a smooth $1$-manifold structure on $X$.

By [L1], the long line $\mathbb{L}$ is Hausdorff and locally Euclidean.

The same cited refutation shows that $\mathbb{L}$ fails second countability, so it is not a manifold under the library convention. The structural details of [F1] identify the witness but do not change that conclusion.

Thus $\mathbb{L}$ is the required counterexample.

Each singleton family $\mathcal{A}$ and $\mathcal{B}$ covers $\mathbb{R}$, so [F1, F2] by [F2] each is an atlas provided its single chart is legitimate, and [F1] supplies that legitimacy.

The transition $\psi \circ {\mathrm{id}}^{-1}(x)=x^3$ is smooth, but the [F3, step 1.1] reverse transition $\mathrm{id}\circ {\psi }^{-1}(x)=x^{1\mathrm{/}3}$ is not differentiable at $0$. Hence [F3] says the two charts are not compatible.

Therefore $\mathcal{A}$ and $\mathcal{B}$ are atlases on the same manifold [step 2.1] that are not compatible, which refutes the statement.

By [L1], the map $F$ is smooth and bijective.

The same cited refutation shows that $F^{-1}$ is not smooth, so [F1] rules [F1, L1] out $F$ being a diffeomorphism.

Hence $F(x)=x^3$ is the desired counterexample.

By [L1], every singleton of $X$ is open and any basis of $X$ must contain [L1] uncountably many distinct singleton sets.

Therefore $X$ is not second countable in the sense of [F1].

So $X$ is the desired counterexample.