Cotangent space at a rational point

The conormal sequence at the point. Apply [F1] to the ring map $k\to R$ and the ideal $\mathfrak{m}\subseteq R$ with quotient $R\mathrm{/}\mathfrak{m}=k$: the sequence $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2⟶k{\otimes }_R{\mathrm{\Omega }}_{R\mathrm{/}k}⟶{\mathrm{\Omega }}_{k\mathrm{/}k}⟶0$ is exact, the first map sending $[a]$ to $1\otimes \mathrm{d}a$, and the middle term is $k{\otimes }_R{\mathrm{\Omega }}_{R\mathrm{/}k}$ with $k=R\mathrm{/}\mathfrak{m}$. By [F2] the last term vanishes, so the first map is surjective.

A retraction. Define $D\text{\hspace{0.17em}⁣}:R\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ by $D(a):=[a-\epsilon (a)]$, where $\epsilon \text{\hspace{0.17em}⁣}:R\to R\mathrm{/}\mathfrak{m}=k$ is the residue map and $[\cdot ]$ is the class modulo ${\mathfrak{m}}^2$. Then $D$ is additive, kills $k$ since $\epsilon$ is the identity on $k\subseteq R$, and is a $k$-derivation: $D(ab)-aD(b)-bD(a)=[-(a-\epsilon (a))(b-\epsilon (b))]=0$ in $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, because both $a-\epsilon (a)$ and $b-\epsilon (b)$ belong to $\mathfrak{m}$. Here elements of $k$ are viewed in $R$ via its structure map, which splits $\epsilon$, and $R$ acts on $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ through $\epsilon$. By [F2] applied to $k\to R$ there is an $R$-linear $\tilde{D}\text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{R\mathrm{/}k}\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ with $\tilde{D}(\mathrm{d}a)=D(a)$; since $\mathfrak{m}\cdot (\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2)=0$, it kills $\mathfrak{m}{\mathrm{\Omega }}_{R\mathrm{/}k}$ and therefore factors through an $R$-linear, hence $k$-linear, map $\psi \text{\hspace{0.17em}⁣}:k{\otimes }_R{\mathrm{\Omega }}_{R\mathrm{/}k}\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$.

The identification of the target. By [F3] applied to an affine chart of $X$ containing $x$, the stalk of ${\mathrm{\Omega }}_{X\mathrm{/}k}$ at $x$ is ${\mathrm{\Omega }}_{R\mathrm{/}k}$, so the relative cotangent space of [F4], namely the residue-field tensor product ${\mathrm{\Omega }}_{X\mathrm{/}k}{\otimes }_{{\mathcal{O}}_{X,x}}\kappa (x)$ of the statement, is $k{\otimes }_R{\mathrm{\Omega }}_{R\mathrm{/}k}$; under this identification the element ${\mathrm{d}}_{X\mathrm{/}k}(a)\otimes 1$ for $a\in R$ corresponds to $1\otimes \mathrm{d}a$. Hence the map of the statement is the first map of the exact sequence of step 1.1, and it is natural in $(X,x)$ because the identification is induced by the universal derivation and localization.

$\psi$ is a left inverse of the first map. For $a\in \mathfrak{m}$ one has $\psi (1\otimes \mathrm{d}a)=\tilde{D}(\mathrm{d}a)=D(a)=[a]$, because $\epsilon (a)=0$. Hence $\psi$ is a left inverse of the map $[a]\mapsto 1\otimes \mathrm{d}a$ of step 1.1, which is therefore injective.

Conclusion. The map of step 1.1 is surjective by step 1.1 and injective by step 2.2, hence an isomorphism $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\cong k{\otimes }_R{\mathrm{\Omega }}_{R\mathrm{/}k}$; by step 2.1 this is exactly the map of the statement, which is therefore an isomorphism of $k$-vector spaces, natural in $(X,x)$. Nothing was used about $x$ beyond $\kappa (x)=k$, and the hypothesis is essential to the argument: for a point with $\kappa (x)\mathrm{\ne }k$ the residue map is not a $k$-algebra section of $R\to \kappa (x)$ in general, and no such retraction is constructed.