The affine atlas makes the Minkowski model a smooth 4-manifold diffeomorphic to R^4

The chart $x$ is a bijection $A\to {\mathbb{R}}^4$ by [L1]. Its preimages of Euclidean open sets form a topology: preimages preserve arbitrary unions and finite intersections and carry $\mathrm{\varnothing },{\mathbb{R}}^4$ to $\mathrm{\varnothing },A$, respectively, so [L3] applies. By the definition of this transported topology $x^{-1}(V)$ is open in $A$ for every open $V\subseteq {\mathbb{R}}^4$ while $(x^{-1}{)}^{-1}(U)=x(U)$ is open in ${\mathbb{R}}^4$ for every open $U\subseteq A$; hence $x$ and $x^{-1}$ are continuous by [L2], so $x$ is a homeomorphism [L2], and the transported topology is the unique topology for which this holds, its open sets being exactly the preimages of the open sets of ${\mathbb{R}}^4$ [L2, L3].

Let $N$ be an invertible real $4\times 4$ matrix and $c\in {\mathbb{R}}^4$, and put $g(u):=Nu+c$. Then $g$ is a bijection of ${\mathbb{R}}^4$ with inverse $g^{-1}(w)=N^{-1}(w-c)$ by [L16]. Each component is $g_l(u)={\sum }_j{N}_{lj}{u}_j+c_l$; by the definition of the partial derivative [L14], ${\mathrm{\partial }}_j{g}_l(a)=N_{lj}$ for every $a$, a constant function, and every iterated partial derivative of order at least $2$ is the partial derivative of a constant, hence exists and equals $0$; therefore each component is of class $C^k$ for every $k$ by [L14], so $g$ is smooth by [L11]. Moreover $g$ and $g^{-1}$ are continuous: $\mid (Nx{)}_l\mid \le ({\max }_j\mid x_j\mid ){\sum }_j\mid N_{lj}\mid \le \parallel x{\parallel }_2{\sum }_j\mid N_{lj}\mid$ by [L15], so $\parallel g(u)-g(v){\parallel }_2=\parallel N(u-v){\parallel }_2\le K_N\parallel u-v{\parallel }_2$ with $K_N:=({\sum }_l({\sum }_j\mid N_{lj}\mid {)}^2{)}^{1\mathrm{/}2}$, which is the Lipschitz condition giving the $\epsilon$-$\delta$ definition of continuity, and the same estimate with $N^{-1}$ in place of $N$ applies to $g^{-1}$.

${\mathbb{R}}^4$ with its standard topology is metrizable by [L7], so distinct points of ${\mathbb{R}}^4$ have disjoint open neighbourhoods by [L6]. For $p\mathrm{\ne }q$ in $A$ the points $x(p)\mathrm{\ne }x(q)$ are distinct because $x$ is injective by step 1.1; choosing disjoint open sets $V\ni x(p)$ and $W\ni x(q)$ in ${\mathbb{R}}^4$, the preimages $x^{-1}(V)$ and $x^{-1}(W)$ are open in $A$ by step 1.1 and are disjoint, so $A$ is Hausdorff by [L5].

The rational open boxes form a countable basis $\mathcal{B}$ of ${\mathbb{R}}^4$ by [L8]. The family $\{x^{-1}(B):B\in \mathcal{B}\}$ is countable and consists of open subsets of $A$ by step 1.1; and it is a basis for the transported topology: if $U\subseteq A$ is open, then $x(U)$ is open in ${\mathbb{R}}^4$ by step 1.1 and equals the union of the members of $\mathcal{B}$ contained in it, so $U$ is the union of the corresponding sets $x^{-1}(B)$, each open in $A$. Hence $A$ is second countable by [L9].

Let $y(p)=Mx(p)+b$ with $M$ invertible and $b\in {\mathbb{R}}^4$, and put $g(u):=Mu+b$. Then $y=g\circ x$ is a bijection $A\to {\mathbb{R}}^4$ by steps 1.1 and 1.2; it is continuous because for open $V\subseteq {\mathbb{R}}^4$ the preimage $y^{-1}(V)=x^{-1}(g^{-1}(V))$ is open in $A$ by steps 1.1 and 1.2; and $y^{-1}$ is continuous because for open $U\subseteq A$ its preimage under $y^{-1}$ is $y(U)=g(x(U))$, open in ${\mathbb{R}}^4$ by steps 1.1 and 1.2. Thus $y$ is a homeomorphism onto ${\mathbb{R}}^4$.

Let $(o^{\mathrm{\prime }};e_0^{\mathrm{\prime }},e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$ be another inertial frame with chart $x^{\mathrm{\prime }}$. By the transition law of [L1], $x^{\mathrm{\prime }}=M^{-1}(x-x(o^{\mathrm{\prime }}))=M^{-1}x+b$ with $b:=-M^{-1}x(o^{\mathrm{\prime }})$, where $M$ is the change-of-basis matrix between the two ordered bases and $M^{-1}$ is invertible because both families are ordered bases; this exhibits $x^{\mathrm{\prime }}$ as an affine chart with invertible linear part $M^{-1}$ in the sense of the affine maps of step 1.2.

By step 2.1 the space $A$ is Hausdorff and by step 2.2 it is second countable. It is locally Euclidean of dimension $4$: for every $p\in A$ the whole space $A$ is an open neighbourhood of $p$ by [L3], and $x:A\to {\mathbb{R}}^4$ is a homeomorphism onto the open subset ${\mathbb{R}}^4\subseteq {\mathbb{R}}^4$ by step 1.1. Hence $A$ is a topological $4$-manifold by [L4], proving claim 1; the homeomorphisms $x$ and $y$ established in steps 1.1 and 2.3 are therefore charts by [L10].

The transition map between the charts $(A,x)$ and $(A,y)$ of step 2.3 is $y\circ x^{-1}=g$, and its inverse transition is $x\circ y^{-1}=g^{-1}$; by step 1.2 both are smooth, so the two charts are smoothly compatible by [L11].

Let $x^{\mathrm{\prime }}$ be another inertial chart, written $x^{\mathrm{\prime }}=g^{\mathrm{\prime }}\circ x$ with $g^{\mathrm{\prime }}$ an affine map of ${\mathbb{R}}^4$ with invertible linear part by step 2.4. For a subset $U\subseteq A$, openness in the topology transported by $x^{\mathrm{\prime }}$ means that $x^{\mathrm{\prime }}(U)=g^{\mathrm{\prime }}(x(U))$ is open in ${\mathbb{R}}^4$, which by step 1.2 (a homeomorphism maps open sets to open sets and reflects them) is equivalent to $x(U)$ being open in ${\mathbb{R}}^4$, that is, to $U$ being open in the topology transported by $x$ by step 1.1. Hence the two transported topologies coincide.

The pair $(A,x)$ is a chart by steps 1.1 and [L10]; the one-chart family $\{(A,x)\}$ covers $A$ and its sole transition $x\circ x^{-1}={\mathrm{i}\mathrm{d}}_{{\mathbb{R}}^4}$ is smooth by step 1.2, so it is a smooth atlas by [L12]; the maximal smooth atlas it generates exists and is unique by [L13], and $A$ with this smooth structure is a smooth $4$-manifold by [L12].

Every affine chart $y=Mx+b$ is smoothly compatible with the chart $(A,x)$ by step 3.2, and every inertial chart is an affine chart by step 2.4; hence each of these charts belongs to the smooth structure generated by the atlas $\{(A,x)\}$ of step 4.1, by the criterion of [L13] that two one-chart atlases generate the same structure exactly when their union is a smooth atlas.

${\mathbb{R}}^4$ is an open subset of itself and carries the standard smooth structure generated by the one-chart atlas $\{({\mathbb{R}}^4,\mathrm{i}\mathrm{d})\}$ by [L17]. With respect to the charts $(A,x)$ of step 4.1 and $({\mathbb{R}}^4,\mathrm{i}\mathrm{d})$, the coordinate representative of $x$ is $\mathrm{i}\mathrm{d}\circ x\circ x^{-1}={\mathrm{i}\mathrm{d}}_{{\mathbb{R}}^4}$ and the representative of $x^{-1}$ is likewise ${\mathrm{i}\mathrm{d}}_{{\mathbb{R}}^4}$; the identity of ${\mathbb{R}}^4$ has components $u_l$ whose partial derivatives are the constants ${\delta }_{lj}$ with all higher iterated partial derivatives $0$, so it is smooth by [L14] and [L11]. Hence $x$ is a smooth bijection with smooth inverse, that is, a diffeomorphism $A\to {\mathbb{R}}^4$ by [L18], proving claim 2.

The smooth structures generated by the one-chart atlases $\{(A,x)\}$ and $\{(A,x^{\mathrm{\prime }})\}$ of two inertial charts coincide: by step 2.4 and step 5.1 the chart $x^{\mathrm{\prime }}$ is smoothly compatible with $x$, so the union of the two one-chart atlases is a smooth atlas and [L13] gives equality of the generated maximal atlases; combined with step 3.3 this proves claim 3 for the transition and claim 4.