Differential of an S-morphism

Construction. By [F1], applied to the $S$-morphism $f$, there is a unique ${\mathcal{O}}_X$-linear $\mathrm{d}f\text{\hspace{0.17em}⁣}:f^{\ast }{\mathrm{\Omega }}_{Y\mathrm{/}S}\to {\mathrm{\Omega }}_{X\mathrm{/}S}$ with $\mathrm{d}f(1\otimes {\mathrm{d}}_{Y\mathrm{/}S}(g))={\mathrm{d}}_{X\mathrm{/}S}(g\circ f)$ for local sections $g$ of ${\mathcal{O}}_Y$; it is obtained by applying the universal property [F4] to the $S$-derivation ${\mathcal{O}}_Y\to f_{\ast }{\mathrm{\Omega }}_{X\mathrm{/}S}$, $g\mapsto {\mathrm{d}}_{X\mathrm{/}S}(g\circ f)$, and then the adjunction of Pullback of modules is left adjoint to pushforward, and it is natural in the data $(X,Y,f)$ by construction.

Identity. For $f={\mathrm{i}\mathrm{d}}_X$ the map sends $1\otimes {\mathrm{d}}_{X\mathrm{/}S}(g)$ to ${\mathrm{d}}_{X\mathrm{/}S}(g)$; since the elements ${\mathrm{d}}_{X\mathrm{/}S}(g)$ generate ${\mathrm{\Omega }}_{X\mathrm{/}S}$ over ${\mathcal{O}}_X$ by [F4], this is the canonical identification ${\mathrm{i}\mathrm{d}}_X^{\ast }{\mathrm{\Omega }}_{X\mathrm{/}S}\cong {\mathrm{\Omega }}_{X\mathrm{/}S}$.

Chain rule. Let $X\stackrel{f}{\to }Y\stackrel{g}{\to }Z$ be morphisms of $S$-schemes. Both $\mathrm{d}(g\circ f)$ and the composite $\mathrm{d}f\circ f^{\ast }(\mathrm{d}g)$ are ${\mathcal{O}}_X$-linear maps $(g\circ f{)}^{\ast }{\mathrm{\Omega }}_{Z\mathrm{/}S}\to {\mathrm{\Omega }}_{X\mathrm{/}S}$ (the composite being formed with the identification [F3]), and on a generator $1\otimes 1\otimes {\mathrm{d}}_{Z\mathrm{/}S}(h)$ both take the value ${\mathrm{d}}_{X\mathrm{/}S}(h\circ g\circ f)$: the composite because $\mathrm{d}f(1\otimes {\mathrm{d}}_{Y\mathrm{/}S}(h\circ g))={\mathrm{d}}_{X\mathrm{/}S}(h\circ g\circ f)$ and $\mathrm{d}g(1\otimes {\mathrm{d}}_{Z\mathrm{/}S}(h))={\mathrm{d}}_{Y\mathrm{/}S}(h\circ g)$, and $\mathrm{d}(g\circ f)$ by its definition. As the generators $1\otimes 1\otimes {\mathrm{d}}_{Z\mathrm{/}S}(h)$ generate the source over ${\mathcal{O}}_X$, the two maps agree.

Fibres and the tangent map. Fix $x\in X$ and put $y=f(x)$. By [F2], the source stalk of $\mathrm{d}f$ is ${\mathrm{\Omega }}_{Y\mathrm{/}S,y}{\otimes }_{{\mathcal{O}}_{Y,y}}{\mathcal{O}}_{X,x}$. Tensoring it with $\kappa (x)$ gives ${\mathrm{\Omega }}_{Y\mathrm{/}S,y}{\otimes }_{{\mathcal{O}}_{Y,y}}\kappa (x)$, canonically $({\mathrm{\Omega }}_{Y\mathrm{/}S,y}{\otimes }_{{\mathcal{O}}_{Y,y}}\kappa (y)){\otimes }_{\kappa (y)}\kappa (x)$, because ${\mathcal{O}}_{Y,y}\to \kappa (x)$ factors through the residue field $\kappa (y)$. Thus the fibre of $\mathrm{d}f$ is the $\kappa (x)$-linear cotangent map displayed in the statement. Dualising over $\kappa (x)$ gives the stated map from $T_{X\mathrm{/}S,x}$ to the $\kappa (x)$-dual of the extended cotangent space at $y$. When the residue-field map is an isomorphism, this target is $T_{Y\mathrm{/}S,y}$; without that hypothesis, the latter is only a $\kappa (y)$-vector space and cannot be the target of a $\kappa (x)$-linear map.

Conclusion. Step 1.1 gives the asserted map and its characterisation, steps 2.1 and 2.2 give the identity and chain rules, and step 2.3 gives the fibre and tangent maps; nothing beyond the universal property of $\mathrm{\Omega }$ and the functoriality of pullback and of extension of scalars was used, so no finiteness, flatness or separatedness hypothesis enters.