Naturality alone gives Mayer–Vietoris connector compatibility

On the unit circle let $U=S^1\setminus \{(1,0)\}$ and $V=S^1\setminus \{(-1,0)\}$. Each is a connected open arc, while $U\cap V$ has two connected open-arc components. By [F1], $H_{\mathrm{d}\mathrm{R}}^0(U)\cong H_{\mathrm{d}\mathrm{R}}^0(V)\cong \mathbb{R}$ and $H_{\mathrm{d}\mathrm{R}}^0(U\cap V)\cong {\mathbb{R}}^2$. The difference of the restrictions of any two constants is diagonal, so the image of $H^0(U)\oplus H^0(V)\to H^0(U\cap V)$ is $\{(c,c):c\in \mathbb{R}\}$. Let $a=(0,1)$. It is not diagonal, and exactness in [F1] therefore gives $\mathrm{\Delta }a\mathrm{\ne }0$ in $H_{\mathrm{d}\mathrm{R}}^1(S^1)$. Now define degreewise maps $T_X^q=(-1{)}^q{\mathrm{id}}_{H_{\mathrm{d}\mathrm{R}}^q(X)}$ for every manifold or open submanifold $X$. For every inclusion $j:X\hookrightarrow Y$, scalar linearity gives $j^{\ast }{T}_Y^q=(-1{)}^q{j}^{\ast }=T_X^q{j}^{\ast }$. Thus $T$ commutes with every ordinary restriction map in every degree.

For the class $a$ of step 1.1, however, $T_{S^1}^1(\mathrm{\Delta }a)=-\mathrm{\Delta }a,\mathrm{\Delta }(T_{U\cap V}^{0}a)=\mathrm{\Delta }a\mathrm{.}$ These values are unequal because $\mathrm{\Delta }a\mathrm{\ne }0$ in a real vector space. Hence restriction naturality alone does not imply connector compatibility.

The actual integration comparison is not the artificial family $T$: [F2] establishes its connector square, including the second-minus-first sign. That theorem is genuinely additional information beyond ordinary restriction naturality, precisely as the counterexample shows. If the overlap, class, or connector is zero, the square may commute vacuously and does not rescue the universal assertion. The counterexample uses $q=0$ and a nonempty disconnected overlap; it has no boundary endpoint or degenerate-chain issue. The displayed counterexample is conditional on the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ branch because [F1] obtains an exact Mayer–Vietoris sequence under that assumption; with a supplied partition the same finite calculation is choice-free by [F2].