Inertial charts are related by proper orthochronous affine Lorentz maps, and every such map is a chart transition

Define $L:V\to V$ by $L(u^0{e}_0+\cdots +u^3{e}_3):=u^0{e}_0^{\mathrm{\prime }}+\cdots +u^3{e}_3^{\mathrm{\prime }}$; by [L3] the coordinate maps $E,E^{\mathrm{\prime }}$ are linear isomorphisms, so $L=E^{\mathrm{\prime }}\circ E^{-1}$ is linear with $L(e_a)=e_a^{\mathrm{\prime }}$, it is unique with this property, and $M$ is its matrix in the ordered basis $(e_a)$ by [L5]. Since both bases are $\eta$-orthonormal by [L3], $\eta (L{e}_a,L{e}_b)=\eta (e_a^{\mathrm{\prime }},e_b^{\mathrm{\prime }})=\eta (e_a,e_b)$ for all $a,b$; expanding two general vectors in the basis and using bilinearity of $\eta$ gives $\eta (Lv,Lw)=\eta (v,w)$ for all $v,w$, so $L$ is a linear Lorentz map; in particular $M^T{\eta }_{0}M={\eta }_0$ by [L1], and $L{e}_0=e_0^{\mathrm{\prime }}\in C^{+}$ by [L3].

By the transition law [L10], $x^{\mathrm{\prime }}(p)=M^{-1}(x(p)-x(o^{\mathrm{\prime }}))=M^{-1}x(p)-M^{-1}x(o^{\mathrm{\prime }})$ for all $p\in A$; under the identification of events with their $x$-coordinates this is the affine map $u\mapsto M^{-1}(u-x(o^{\mathrm{\prime }}))$ of ${\mathbb{R}}^4$, with linear part $M^{-1}$ and constant part $-M^{-1}x(o^{\mathrm{\prime }})$.

Let $L$ be a proper orthochronous linear Lorentz map and $c\in A$. The images $L{e}_0,\dots ,L{e}_3$ form an ordered basis, and $\eta (L{e}_a,L{e}_b)=\eta (e_a,e_b)=\mathrm{diag}(-1,1,1,1{)}_{ab}$, so it is $\eta$-orthonormal; $L{e}_0\in C^{+}$ because $e_0\in C^{+}$ by [L3] and $L$ is orthochronous; and it is positively oriented, because the change-of-basis map from $(e_a)$ to $(L{e}_a)$ is $L$ itself with $\det L=+1>0$, so both ordered bases lie in one class by [L4]. Hence $(c;L{e}_0,L{e}_1,L{e}_2,L{e}_3)$ is an inertial frame. Its coordinate isomorphism $E_L(u):={\sum }_a{u}^{a}L{e}_a$ satisfies $E_L(u)=L\left({\sum }_a{u}^a{e}_a\right)=L(E(u))$ by linearity, so its chart is $x_L(p)=E_L^{-1}(p-c)=E^{-1}(L^{-1}(p-c))=M_L^{-1}(x(p)-x(c))$, with $M_L=E^{-1}\circ L\circ E$ and $M_L^{-1}=E^{-1}\circ L^{-1}\circ E$, so $x_L(p)=M_L^{-1}(x(p)-x(c))$ for all $p\in A$.

By step 1.1 the matrix $M$ is the matrix of the linear Lorentz map $L$, so $\det L=\det M$ by [L6]; by [L2], $\det L=\pm 1$. Both $(e_a)$ and $(e_a^{\mathrm{\prime }})$ are positively oriented by [L3], hence lie in the same orientation class of [L4], so the change-of-basis map $P$ with $P(e_a)=e_a^{\mathrm{\prime }}$ satisfies $\det P>0$; by the uniqueness in step 1.1, $P=L$, so $\det M=\det L>0$ and therefore $\det M=+1$.

By step 1.1, $L{e}_0=e_0^{\mathrm{\prime }}\in C^{+}$ lies in $L(C^{+})$, so $L(C^{+})\cap C^{+}\mathrm{\ne }\mathrm{\varnothing }$; by [L2] the set $L(C^{+})$ is one of the two components $C^{+},C^{-}$, hence it cannot be $C^{-}$ and $L(C^{+})=C^{+}$.

By [L2] the inverse $L^{-1}$ is a linear Lorentz map, and its matrix in the ordered basis $(e_a)$ is $M^{-1}$ by [L5]; moreover $L^{-1}$ is proper by the subgroup assertion of [L2], since $L$ is proper orthochronous by steps 2.1 and 2.2, and $L^{-1}(C^{+})=C^{+}$ because $L(C^{+})=C^{+}$ from step 2.2 and $L$ is a bijection. Hence $M^{-1}$ is the matrix of the proper orthochronous Lorentz map $L^{-1}$, and $M^{-1}$ maps the future cone onto the future cone.

Let $A:=M^{-1}$ and $u$ belong to the future closed coordinate cone, that is $q(u)\le 0$ and $u^0\ge 0$, which by [L8] and [L7] is exactly the condition that $E(u)$ lies in the closed future cone $\mathrm{cl}(C^{+})$. Put $w:=Au$. First, $q(w)={\eta }_0(Au,Au)={\eta }_0(u,u)=q(u)\le 0$, because $A$ is the matrix of the Lorentz map $L^{-1}$ by step 3.1 and $E(M^{-1}u)=L^{-1}(E(u))$ by step 1.1's construction $L\circ E=E\circ M$. Second, define the coordinate column ${\epsilon }_0:=(1,0,0,0)\in {\mathbb{R}}^4$, so $E({\epsilon }_0)=e_0$, and put $z:=A^{-1}{\epsilon }_0=M{\epsilon }_0\in {\mathbb{R}}^4$. Then $E(z)=L(E({\epsilon }_0))=L{e}_0=e_0^{\mathrm{\prime }}\in C^{+}$ by step 1.1 and [L3], so $z^0>\mathrm{\mid }{z}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }\ge 0$ by [L8]. Since $A^T{\eta }_{0}A={\eta }_0$ implies ${\eta }_0(Au,v)={\eta }_0(u,A^{-1}v)$ for all coordinate columns $u,v$, we get $w^0=-{\eta }_0(Au,{\epsilon }_0)=-{\eta }_0(u,z)=u^0{z}^0-u_{\mathrm{v}\mathrm{e}\mathrm{c}}\cdot z_{\mathrm{v}\mathrm{e}\mathrm{c}}\ge u^0{z}^0-\mathrm{\mid }{u}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }\mathrm{\mid }{z}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }\ge u^0(z^0-\mathrm{\mid }{z}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid })\ge 0$, where the middle step is [L9] and the last uses $u^0\ge \mathrm{\mid }{u}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }$. Hence $q(w)\le 0$ and $w^0\ge 0$, so $w$ lies in the future closed coordinate cone. If $u$ is future causal, it is additionally nonzero; invertibility gives $w\mathrm{\ne }0$, so $w$ is future causal as well. The linear map $A$ fixes zero, which is not called causal. Replacing $A$ by $A^{-1}=M$ gives the reverse cone inclusion by the same argument, since $M$ is also proper orthochronous by steps 1.1, 2.1 and 2.2. For the affine transition $F$ of step 1.2, translation cancels: $F(u)-F(v)=A(u-v)$ for all $u,v\in {\mathbb{R}}^4$. Applying the identities just proved to $u-v$ gives $q(F(u)-F(v))=q(u-v)$ and preservation of future causal displacements.

Clause (i) is step 1.1 with the properness of step 2.1, the orthochrony of step 2.2 and the statements about $M^{-1}$ in step 3.1. Clause (ii) is the transition formula of step 1.2 together with the preservation of $q$ on differences and of future causal displacements in step 4.1. Clause (iii) is step 1.3, and conversely every transition arises there: given the two frames, step 1.1 produces the proper orthochronous Lorentz map $L$ with $L{e}_a=e_a^{\mathrm{\prime }}$, and step 1.2 writes the transition as $u\mapsto M^{-1}(u-x(o^{\mathrm{\prime }}))=M_L^{-1}(u-x(c))$ with $M_L=M$ and $c:=o^{\mathrm{\prime }}$, so the transitions from $x$ are exactly the maps listed, and their linear parts are exactly the proper orthochronous Lorentz matrices and their inverses because step 3.1 computes the linear part of the reverse transition as $M$ for the same pair.