The oriented volume form is independent of the positively oriented inertial frame

From $e_a^{\mathrm{\prime }}={\sum }_b{M}_a^b{e}_b$ and the $\eta$-orthonormality of both frames [L2, L3], ${\eta }_{ab}=\eta (e_a^{\mathrm{\prime }},e_b^{\mathrm{\prime }})={\sum }_{c,d}{M}_a^c{M}_b^d\eta (e_c,e_d)=(M^{\mathsf{T}}\eta M{)}_{ab}$, that is, $M^{\mathsf{T}}\eta M=\eta$ with $\eta =\mathrm{diag}(-1,1,1,1)$; the Leibniz formula of For $n\ge 1$, the determinant over a commutative ring by the Leibniz formula, and $\mathrm{\mid }\det A\mathrm{\mid }$ for a real matrix gives $\det (M^{\mathsf{T}})=\det M$ by replacing each permutation by its inverse (which has the same sign) and reindexing its finite product. Taking determinants and using [L6] therefore gives $(\det M{)}^2\det \eta =\det \eta$, and $\det \eta =-1$, so $(\det M{)}^2=1$.

Both ordered bases lie in the positive class selected by the fixed orientation of $V$ [L2], so the change of basis between them has determinant $>0$ by [L4]; together with $(\det M{)}^2=1$ from step 1.1 this gives $\det M=+1$.

Let $X=(X_j^b)$ and $X^{\mathrm{\prime }}=(X_j^{\mathrm{\prime }b})$ be the coordinate matrices of a tuple $(v_0,v_1,v_2,v_3)$ in the two frames. Substituting $e_a^{\mathrm{\prime }}={\sum }_b{M}_a^b{e}_b$ into $v_j={\sum }_a{X}_j^{\mathrm{\prime }a}{e}_a^{\mathrm{\prime }}={\sum }_a{X}_j^a{e}_a$ gives $X_j^b={\sum }_a{M}_a^b{X}_j^{\mathrm{\prime }a}$, that is, $X=M{X}^{\mathrm{\prime }}$, hence $X^{\mathrm{\prime }}=M^{-1}X$ and, by [L6] and [L7], $\det X^{\mathrm{\prime }}=\det (M^{-1})\det X=(\det M{)}^{-1}\det X=\det X$ because $\det M=1$ by step 2.1.

By step 3.1 the two determinants that [L1] uses to define $ϵ(v_0,v_1,v_2,v_3)$ in the two frames agree, so $ϵ$ does not depend on the inertial frame; in particular the coordinate matrix of the frame $(e_0,e_1,e_2,e_3)$ in the frame $(e_0^{\mathrm{\prime }},e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$ is $M^{-1}$, so $ϵ(e_0,e_1,e_2,e_3)=\det (M^{-1})=1$ by [L7] and step 2.1. The same determinant argument applies to any positively oriented $\eta$-orthonormal ordered basis, including one with past time axis: steps 1.1 through 3.1 use orthonormality and positive orientation, but not the future condition. Hence $ϵ$ has the asserted normalization on every such basis.

The raised component: in an inertial frame [L3] gives ${\eta }^{aa}=\mathrm{diag}(-1,1,1,1)$ and ${ϵ}_{0123}=1$ by [L1] and step 4.1, so ${ϵ}^{0123}=(-1)(+1)(+1)(+1)\cdot 1=-1$; since $ϵ$ is frame independent by step 4.1, this value is the same in every inertial frame.

For uniqueness, let ${ϵ}^{\mathrm{\prime }}$ be any alternating $4$-linear form on $V$ with ${ϵ}^{\mathrm{\prime }}(f_0,f_1,f_2,f_3)=1$ for every positively oriented $\eta$-orthonormal ordered basis. Fix an inertial frame $(e_a)$ and a tuple $(v_0,v_1,v_2,v_3)$ with coordinate matrix $X$. Expanding each $v_j={\sum }_a{X}_j^a{e}_a$ and using multilinearity and alternation [L5] together with the normalization and [L8], ${ϵ}^{\mathrm{\prime }}(v_0,v_1,v_2,v_3)=\det (X){ϵ}^{\mathrm{\prime }}(e_0,e_1,e_2,e_3)=\det X$, which equals $ϵ(v_0,v_1,v_2,v_3)$ by [L1]. Hence $ϵ$ is the unique alternating $4$-linear form normalized to $1$ on every positively oriented $\eta$-orthonormal ordered basis, which completes the proof of all the displayed claims.