The metric pairing is a linear bijection to the dual, and the raised covector in an orthonormal frame

The map $\mathrm{\Phi }$ is linear, and $\ker \mathrm{\Phi }=\{0\}$. Since $\eta$ is a bilinear form on $V$, its associated map ${\eta }^{\mathrm{♭}}$ lies in $\mathcal{L}(V,V^{\ast })$ and ${\eta }^{\mathrm{♭}}(v)(w)=\eta (v,w)=\mathrm{\Phi }(v)(w)$ for all $v,w$, so $\mathrm{\Phi }={\eta }^{\mathrm{♭}}$ is linear by [L2]. If $\mathrm{\Phi }(v)=0$, then $\eta (v,w)=0$ for every $w\in V$, so $v$ lies in the left radical of $\eta$, which is $\{0\}$ by [L1]; conversely $\mathrm{\Phi }(0)=0$ by linearity (equivalently $\eta (0,\cdot )=0$), so $\ker \mathrm{\Phi }=\{0\}$.

The map $\mathrm{\Phi }$ is surjective. Fix an $\eta$-orthonormal ordered basis $(e_0,e_1,e_2,e_3)$ of $V$ [L4], with dual family $(e^0,e^1,e^2,e^3)$, a basis of $V^{\ast }$ by [L3]. For all $a,b$ one has $\mathrm{\Phi }(e_a)(e_b)=\eta (e_a,e_b)={\eta }_{aa}{\delta }_{ab}$ with ${\eta }_{aa}\in \{-1,1\}$, by [L4], and the functional ${\eta }_{aa}{e}^a$ takes the same values on the basis $(e_b)$; two linear functionals that agree on a basis agree on every vector, so $\mathrm{\Phi }(e_a)={\eta }_{aa}{e}^a$, which exhibits $e^a={\eta }_{aa}\mathrm{\Phi }(e_a)$ as a scalar multiple of $\mathrm{\Phi }(e_a)$ (here ${\eta }_{aa}^{-1}={\eta }_{aa}$). Hence every element of the basis $(e^a)$ of $V^{\ast }$ lies in the linear span of $\mathrm{\Phi }(e_0),\dots ,\mathrm{\Phi }(e_3)$, so that span is all of $V^{\ast }$; therefore $\mathrm{\Phi }$ is surjective.

The map $\mathrm{\Phi }$ is a linear bijection, $\dim V^{\ast }=\dim V=4$, and the inverse assignment $\mathrm{♯}$ exists with the stated properties. By step 1.1 $\mathrm{\Phi }$ is injective and by step 1.2 it is surjective, hence bijective; by [L3] and [L1], $\dim V^{\ast }=\dim V=4$; and [L5] applied to the linear map $\mathrm{\Phi }$ gives the same picture independently, $\mathrm{nullity}\mathrm{\Phi }=\dim \ker \mathrm{\Phi }=0$ and $\mathrm{rank}\mathrm{\Phi }=4=\dim V^{\ast }$, with $\mathrm{rank}\mathrm{\Phi }=\dim \mathrm{im}\mathrm{\Phi }$ and $\mathrm{im}\mathrm{\Phi }=V^{\ast }$ by step 1.2. By [L6] the inverse ${\mathrm{\Phi }}^{-1}:V^{\ast }\to V$ is a well-defined linear map, unique, and satisfies ${\mathrm{\Phi }}^{-1}\circ \mathrm{\Phi }={\mathrm{id}}_V$ and $\mathrm{\Phi }\circ {\mathrm{\Phi }}^{-1}={\mathrm{id}}_{V^{\ast }}$. Writing $k^{\mathrm{♯}}:={\mathrm{\Phi }}^{-1}(k)$ for $k\in V^{\ast }$, we get $\eta (k^{\mathrm{♯}},w)=\mathrm{\Phi }(k^{\mathrm{♯}})(w)=k(w)$ for every $w\in V$; any $c$ with $\eta (c,w)=k(w)$ for all $w$ satisfies $\mathrm{\Phi }(c)=k$, hence $c=k^{\mathrm{♯}}$ by injectivity of $\mathrm{\Phi }$, which is the asserted uniqueness. The two inverse equations are exactly $(v^{\mathrm{♭}}{)}^{\mathrm{♯}}=v$ for $v^{\mathrm{♭}}=\mathrm{\Phi }(v)$ and $(k^{\mathrm{♯}}{)}^{\mathrm{♭}}=k$, and $k^{\mathrm{♯}}=0$ holds if and only if $k=\mathrm{\Phi }(k^{\mathrm{♯}})=\mathrm{\Phi }(0)=0$. This proves (a), (b) and (c).

Component formula. Let $k\in V^{\ast }$ and keep the orthonormal basis $(e_a)$ of step 1.2; put $c^0:=-k_0$ and $c^i:=k_i$ for $i=1,2,3$, where $k_a=k(e_a)$, and set $c:=c^0{e}_0+c^1{e}_1+c^2{e}_2+c^3{e}_3$. For every basis vector $e_b$, bilinearity and orthonormality [L1, L4] give $\eta (c,e_b)={\sum }_a{c}^a\eta (e_a,e_b)=c^b{\eta }_{bb}$; hence $\eta (c,e_0)=-c^0=k_0=k(e_0)$ and $\eta (c,e_i)=c^i=k_i=k(e_i)$ for $i=1,2,3$. The functionals $w\mapsto \eta (c,w)$ (linear in its second argument by [L1]) and $k$ agree on the basis $(e_b)$, hence on all of $V$; therefore $\mathrm{\Phi }(c)=k$, and uniqueness in step 2.1 yields $c=k^{\mathrm{♯}}=-k_0{e}_0+k_1{e}_1+k_2{e}_2+k_3{e}_3$.

Quadratic form and causal trichotomy. With $k^{\mathrm{♯}}={\sum }_a{c}^a{e}_a$ from step 3.1, bilinearity and orthonormality give $\eta (k^{\mathrm{♯}},k^{\mathrm{♯}})={\sum }_{a,b}{c}^a{c}^b\eta (e_a,e_b)={\sum }_a(c^a{)}^2{\eta }_{aa}=-(c^0{)}^2+(c^1{)}^2+(c^2{)}^2+(c^3{)}^2=-(k_0{)}^2+(k_1{)}^2+(k_2{)}^2+(k_3{)}^2$, because $(c^0{)}^2=(k_0{)}^2$ and $c^i=k_i$. If $k\mathrm{\ne }0$ then $k^{\mathrm{♯}}\mathrm{\ne }0$ by step 2.1, so [L7] applies: $k^{\mathrm{♯}}$ is null exactly when this value is $0$, that is when $(k_0{)}^2=(k_1{)}^2+(k_2{)}^2+(k_3{)}^2$; timelike exactly when it is negative, that is when $(k_0{)}^2>(k_1{)}^2+(k_2{)}^2+(k_3{)}^2$; and spacelike exactly when it is positive, that is when $(k_0{)}^2<(k_1{)}^2+(k_2{)}^2+(k_3{)}^2$. If $k=0$ then $k^{\mathrm{♯}}=0$ by step 2.1 and the zero vector has no causal character by [L7]. No orientation, time orientation or affine structure was used, and no choice principle was used.