Cotangent spaces commute with localization at a rational point

Localization of the powers. If $ab\in \mathfrak{m}$, then $\bar{a}\bar{b}=0$ in the field $A\mathrm{/}\mathfrak{m}$, so $\bar{a}=0$ or $\bar{b}=0$; hence $\mathfrak{m}$ is prime and $S=A\setminus \mathfrak{m}$ is multiplicative. By [F2] and [F3], $A_{\mathfrak{m}}=S^{-1}A$ and its maximal ideal is $\mathfrak{n}$. By [F4], $\mathfrak{n}=S^{-1}\mathfrak{m}$ and each extension $S^{-1}I$ consists of fractions with numerator in $I$. With the stated recursive convention for powers, [F6] gives ${\mathfrak{n}}^r=S^{-1}({\mathfrak{m}}^r)$ for every $r\ge 0$: it holds for $r=0$. If it holds at $r$, every element of ${\mathfrak{n}}^{r+1}={\mathfrak{n}}^r\mathfrak{n}$ is a finite sum of products $(a_i\mathrm{/}{s}_i)(b_i\mathrm{/}{t}_i)=a_i{b}_i\mathrm{/}(s_i{t}_i)$ with $a_i\in {\mathfrak{m}}^r$ and $b_i\in \mathfrak{m}$, so it lies in $S^{-1}({\mathfrak{m}}^{r+1})$. Conversely, writing a numerator in ${\mathfrak{m}}^{r+1}={\mathfrak{m}}^r\mathfrak{m}$ as a finite sum of such products expresses every fraction in $S^{-1}({\mathfrak{m}}^{r+1})$ as an element of ${\mathfrak{n}}^{r+1}$. Therefore ${\theta }_r$ is well-defined.

Injectivity. Suppose $a\in {\mathfrak{m}}^r$ and ${\theta }_r([a])=0$. By step 1.1 and [F4], $a\mathrm{/}1=b\mathrm{/}s$ for some $b\in {\mathfrak{m}}^{r+1}$ and $s\in S$. By [F5], there is $u\in S$ with $u(as-b)=0$, so $usa=ub\in {\mathfrak{m}}^{r+1}$. The residue of $us$ in the field $A\mathrm{/}\mathfrak{m}$ is nonzero; choose $t\in A$ whose residue is its inverse. Then $v=tus-1\in \mathfrak{m}$ and $a=tusa-va\in {\mathfrak{m}}^{r+1}+\mathfrak{m}{\mathfrak{m}}^r={\mathfrak{m}}^{r+1}$. Thus $[a]=0$ and ${\theta }_r$ is injective.

Surjectivity and boundary instances. Let a class in ${\mathfrak{n}}^r\mathrm{/}{\mathfrak{n}}^{r+1}$ be represented, by step 1.1 and [F4], by $a\mathrm{/}s$ with $a\in {\mathfrak{m}}^r$ and $s\in S$. Choose $t\in A$ whose residue is the inverse of the nonzero residue of $s$. Then $1-ts\in \mathfrak{m}$, and $(a\mathrm{/}s)-(ta\mathrm{/}1)=((1-ts)a)\mathrm{/}s\in S^{-1}({\mathfrak{m}}^{r+1})={\mathfrak{n}}^{r+1}$. Hence the class is ${\theta }_r([ta])$, proving surjectivity. This covers $r=0$, where the map is $A\mathrm{/}\mathfrak{m}\to A_{\mathfrak{m}}\mathrm{/}\mathfrak{n}$, and $r=1$, the cotangent-space map of [F1]. If $\mathfrak{m}=0$, then $A=k$: ${\theta }_0$ is the identity of $k$ and for every $r\ge 1$ both sides are zero. The lifts $t$ above are chosen separately for each displayed fraction, so no choice principle is used.