The tangent space at the identity is a vector space, and Lie is a functor

Classification over an arbitrary $R$. Put $B={\mathcal{O}}_{G,e}$ and $\mathfrak{m}=\ker (B\to k)$. Every morphism $\mathrm{Spec}R[\epsilon ]\to G$ reducing to the constant identity has underlying image $e$, since the nilpotent thickening has the same points as $\mathrm{Spec}R$; it therefore factors through every affine neighbourhood of $e$. Every element of a chart algebra outside the prime of $e$ maps to a unit: its reduction is a nonzero scalar in $k$, and $c+\epsilon r$ has inverse $c^{-1}-\epsilon c^{-2}r$. Consequently such morphisms correspond exactly to $k$-algebra maps $B\to R[\epsilon ]$ of the form $b\mapsto ϵ(b)+\epsilon D(b)$, where $ϵ:B\to k\to R$ is augmentation and $D:B\to R$ is $k$-linear with $D(b{b}^{\mathrm{\prime }})=ϵ(b)D(b^{\mathrm{\prime }})+ϵ(b^{\mathrm{\prime }})D(b)$. The splitting $B=k\oplus \mathfrak{m}$ shows that these $D$ correspond exactly to $k$-linear maps $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\to R$: the product rule kills ${\mathfrak{m}}^2$, and conversely that rule follows by multiplying the two decompositions into scalar and augmentation parts. This gives a natural bijection ${\beta }_R:\mathrm{Lie}(G)(R)\to {\mathrm{Hom}}_k(\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2,R)$, valid also for $R=0$. For $R=k$ it is the coefficient bijection of [F1] and [F5].

Finite-dimensional tensor identification. Write $V=\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$. It is finite-dimensional by [F6], and a basis is obtained by finite elimination as in [F7]. In that basis the natural map $V^{\vee }{\otimes }_{k}R\to {\mathrm{Hom}}_k(V,R)$, $\theta \otimes r\mapsto (v\mapsto r\theta (v))$, identifies both sides with $R^{{\dim }_{k}V}$ and is an $R$-linear isomorphism. Since $V^{\vee }=\mathfrak{g}$ by [F2], transporting this module structure through step 1.1 gives a natural $R$-module structure and a natural identification $\mathrm{Lie}(G)(R)\cong \mathfrak{g}{\otimes }_{k}R$. The pullback of the cotangent sheaf along the identity section is $V{\otimes }_{k}R$ by [F4] and [F5]; this uses a section pullback, not a local ring at a purported general $R$-point.

Multiplication is addition. Apply step 1.1 to $G{\times }_{k}G$ at $(e,e)$. A dual-number morphism to this product is a pair of such morphisms to $G$, so the cotangent fibre of the product is $V\oplus V$: this also follows from the universal product derivation formula $d(a\otimes b)=bda+adb$ on affine charts, followed by augmentation. The cotangent map induced by multiplication is $V\to V\oplus V$ and has both components the identity, since multiplication restricted to $(\mathrm{id},e)$ and $(e,\mathrm{id})$ is the identity by [F3]. Under the coefficient classification of step 1.1, composing two lifts with multiplication therefore takes the pair $(D,D^{\mathrm{\prime }})$ to $D+D^{\mathrm{\prime }}$. Thus the group product on the kernel is precisely addition in the module of step 2.1, for every $R$; it is abelian, its identity is the zero coefficient, and inversion negates the coefficient.

Scalars and the field case. For $c\in R$, the endomorphism $R[\epsilon ]\to R[\epsilon ]$ sending $\epsilon \mapsto c\epsilon$ takes the coefficient $D$ of step 1.1 to $cD$. These operations are the scalar multiplication of the $R$-module of step 2.1 and are natural under $k$-algebra maps $R\to R^{\mathrm{\prime }}$. Taking $R=k$, the dual-number bijection and the cotangent description are isomorphisms of finite-dimensional $k$-vector spaces, as asserted in (a).

Functoriality and closed immersions. A group-scheme morphism $f:G\to H$ preserves identities, so it induces a local homomorphism ${\mathcal{O}}_{H,e_H}\to {\mathcal{O}}_{G,e_G}$ and the corresponding $k$-linear cotangent map $V_H\to V_G$ by [F9]. Precomposition with this map carries a coefficient $D:V_G\to R$ to the coefficient of $f\circ e^{\epsilon X}$ by step 1.1; it is $R$-linear and, under step 2.1, is the scalar extension of its $k$-linear dual $\mathrm{Lie}(f)$. Identity and composite maps give the stated functorial equalities, and $f_R(e^{\epsilon X})=e^{\epsilon \mathrm{Lie}(f{)}_R(X)}$. If $f$ is a closed immersion, [F10] makes its map on $R[\epsilon ]$-points injective for every $R$, hence also on these kernels; taking $R=k$ proves injectivity of $\mathrm{Lie}(f)$. All basis selections are finite, so no choice principle is used.