Differentials, open restriction, and the chain rule

Put $B={\mathcal{O}}_{Y,y}$, $A={\mathcal{O}}_{X,x}$, $\mathfrak{n}={\mathfrak{m}}_y$, and $\mathfrak{m}={\mathfrak{m}}_x$. By [F1], $f_x^{\mathrm{♯}}:B\to A$ is local; [F2] means that $f_x^{\mathrm{♯}}(\mathfrak{n})\subseteq \mathfrak{m}$ and $f_x^{\mathrm{♯}}({\mathfrak{n}}^2)\subseteq {\mathfrak{m}}^2$. It therefore induces a map $\mathfrak{n}\mathrm{/}{\mathfrak{n}}^2\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$. Since $f$ is a $k$-morphism, [F3] says that its stalk map commutes with the two structure maps from $k$; because $x$ and $y$ are rational, these maps identify both residue fields with $k$. The quotient map is thus $k$-linear by [F4]. Dualizing it over $k$ gives the stated map $d_{x}f:T_{x}X\to T_{y}Y$ by [F5].

Let $j:U\to X$ be a $k$-open immersion and let $u\in U(k)$ map to $x\in X(k)$. By [F7], $j$ identifies $U$ with an open subscheme of $X$; by [F8] that open subscheme carries the restricted structure sheaf. Hence the induced stalk map $j_u^{\mathrm{♯}}:{\mathcal{O}}_{X,x}\to {\mathcal{O}}_{U,u}$ is an isomorphism. It identifies maximal ideals and their squares, so the induced cotangent map $C_{x}X\to C_{u}U$ is an isomorphism. Its dual $d_{u}j$ is therefore an isomorphism $T_{u}U\to T_{x}X$.

For the identity morphism, the local-ring and cotangent maps are identities, so their dual is the identity. If $g:Y\to Z$ is another $k$-morphism and $z=g(y)$, contravariance on stalks gives $(g\circ f{)}_x^{\mathrm{♯}}=f_x^{\mathrm{♯}}\circ g_y^{\mathrm{♯}}$. Passing to maximal ideals modulo squares gives ${\overline{(g\circ f)}}_x^{\mathrm{♯}}={\bar{f}}_x^{\mathrm{♯}}\circ {\bar{g}}_y^{\mathrm{♯}}$. Dualizing reverses this order, so $d_x(g\circ f)=({\bar{g}}_y^{\mathrm{♯}}{)}^{\ast }\circ ({\bar{f}}_x^{\mathrm{♯}}{)}^{\ast }=d_{y}g\circ d_{x}f$. This proves identity and chain rules without choosing coordinates or bases.

Under [F6], a tangent vector $t\in T_{x}X$ is represented by a based map ${\gamma }_t:\mathrm{Spec}(k[ϵ]\mathrm{/}({ϵ}^2))\to X$. For $b\in \mathfrak{n}$, the coefficient of $ϵ$ in the pullback of $b$ by $f\circ {\gamma }_t$ is the value of $t$ on $f_x^{\mathrm{♯}}(b)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>{\mathfrak{m}}^2$, namely $t({\bar{f}}_x^{\mathrm{♯}}(b<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>{\mathfrak{n}}^2))$. This is exactly the functional $d_{x}f(t)$ on $C_{y}Y$. Constants have zero $ϵ$-coefficient, so the agreement holds on the whole local ring. Thus the dualized construction is the map on based dual-number points induced by post-composition with $f$; the identity and composition laws also agree with composition of these maps.

If a source or target cotangent space is zero, the induced cotangent map still has the displayed source and target, and its dual is the unique corresponding linear map; in particular zero tangent vectors map to zero. If both cotangent spaces are one-dimensional and the cotangent map sends a chosen target generator to $c$ times a chosen source generator, its dual sends a source functional with value $a$ on the source generator to the target functional with value $ca$ on the target generator. This is precisely the same formula as step 1.1 and introduces no exceptional one-dimensional case. The construction is defined for every local map, including zero, noninjective, or nonsurjective cotangent maps. If $X$ is empty there is no source rational point and the pointwise assertions are vacuous. The zero tangent vector is the based map factoring through $\mathrm{Spec}k$ and is preserved by post-composition. All maps used are canonical, so no choice of bases or other choices, and no Axiom of Choice, is used. The statement contains no iff claim.