Smooth continuous singular cohomology comparison is an isomorphism on convex coordinate domains

If $B$ is nonempty fix one $b\in B$ and let $H(x,t)=(1-t)x+tb$. Convexity makes this a homotopy in $B$ from identity to the constant map. It is smooth, also in half-space coordinates. Transporting through the coordinate diffeomorphism gives a smooth contraction of $W$ to its chosen point. This uses a single point, not a choice indexed by all domains.

By [F3], flattening time gives a smooth-chain prism $P$ from the identity to the constant-map chain operator. Dual precomposition $K^k\phi =\phi P_{k-1}$ gives the cochain homotopy equation $c^{\ast }-1=\delta K+K\delta$, with $K^0=0$. Thus on smooth cohomology the maps induced by the point inclusion $e:\{\ast \}\to W$ and projection $p:W\to \{\ast \}$ are inverse, since $pe=1$ and $ep$ is the contracted constant map.

On ordinary cohomology the same $e,p$ induce inverse maps by [F1]. The squares in [F4] commute with these point maps, and restriction on the point complex is the identity. Therefore $H({\rho }_W)$ is an isomorphism. On a point the unnormalized cochain differential is zero in even degree and identity in odd degree, so its cohomology is $\mathbb{R}$ only in degree zero. This gives the asserted groups for $W$.

If $B$ is empty all complexes and maps are zero. When $n=0$ the nonempty convex domain is one point and the same identity applies. Constant and degenerate simplices are retained throughout; step 2.1 uses the full signed prism, not a normalized quotient. In half-space charts raw time could leave the target beyond an endpoint, which is exactly why step 2.1 uses the flattened target-valued prism. Negative degrees vanish, and no AC is used.