Smooth singular simplices in a coordinate ball

The affine map $L:A^k\to {\mathbb{R}}^n$, $L(\lambda )={\sum }_i{\lambda }_i{v}_i$, is smooth. Convexity implies $L({\mathrm{\Delta }}^k)\subset B$. Therefore $O=L^{-1}(B)$ is open in $A^k$ and contains the entire simplex. The smooth map ${\psi }^{-1}\circ L:O\to W\subset M$ is the extension required by [F1]. This constructs one common neighbourhood directly.

To make the uniform margin explicit, $K=L({\mathrm{\Delta }}^k)$ is compact. If $B=B(b,R)$, the continuous function $y\mapsto \mathrm{\parallel }y-b\mathrm{\parallel }$ attains a maximum $r<R$ on $K$. Put $\epsilon =(R-r)\mathrm{/}2>0$. Any point within distance $\epsilon$ of $K$ lies in $B$ by the triangle inequality. Hence $L^{-1}(\{y:\mathrm{dist}(y,K)<\epsilon \})$ is a single open neighbourhood of the closed simplex on which the same extension is defined. For example, with $B=(-2,2)$ and vertices $-1,1$, the path is $\sigma (t)=2t-1$, whose extension remains in $B$ for $-1\mathrm{/}2<t<3\mathrm{/}2$.

For $k=0$ this gives the constant extension on the one-point affine space. For repeated or coincident vertices the affine formula is still smooth and may be constant; no independence is needed. The construction includes all faces and endpoints because $O$ contains the closed simplex. Empty balls cannot carry the given vertex tuple; in dimension zero the ball is a point and the same formula is constant. Only a supplied finite tuple is used, so no AC is required.