Integral of a form over a smooth singular simplex

For $k=1$, $T_1=[0,1]$ is a compact Jordan interval. If $T_{k-1}$ is compact Jordan for $k\ge 2$, then $T_k$ is its solid under the nonnegative continuous function $u\mapsto 1-{\sum }_{j=1}^{k-1}{u}_j$. Applying [F5] inductively proves compactness and Jordan measurability in every positive dimension. This finite induction chooses no family of objects.

The inverse image under $\bar{\sigma }$ of the open domain of $\omega$ is an open neighbourhood of ${\mathrm{\Delta }}^k$, so the restriction used in the definition exists. By [F3] and [F4], $f$ is smooth on an open neighbourhood of $T_k$, hence continuous on $T_k$. Step 1.1 and [F6] give a finite Riemann integral, and [F7] makes its value independent of a bounding rectangle. The coefficient itself need not have compact support on its neighbourhood.

Two extensions agree on the relative interior of ${\mathrm{\Delta }}^k$, an open set in its affine span; their derivatives and hence pullback coefficients agree there. For any $x\in T_k$, the points $(1-t)x+tb$, where $b_j=1\mathrm{/}(k+1)$ and $0<t\le 1$, lie in the interior and converge to $x$ as $t\downarrow 0$. Continuity of both coefficients forces equality at $x$. Thus their integrands are identical on all of $T_k$, proving extension independence, including along every face.

When $k=0$, evaluation needs no derivative or integration theorem in dimension zero. When $M$ is empty there is no simplex to which the definition applies. A zero form has zero coefficient and integral zero. Degenerate maps are allowed; if the derivative has rank less than $k>0$, alternation makes its top-form pullback zero. The affine domain and its faces are retained even for such maps. The definition uses one extension whose existence is part of [F2], and the resulting value is independent of it; it makes no simultaneous selection of extensions and uses no choice axiom.