Compactly supported cohomology is contravariant for every smooth map

Take the bump $\rho$ from [F2], with $\rho (0)=1$. Its support is closed and bounded, hence compact by [F3], so $\rho \in {\mathrm{\Omega }}_c^0(\mathbb{R})$. Pullback in degree zero is composition, and therefore $F^{\ast }\rho =\rho \circ F=1$ on all of $\mathbb{R}$.

The support of the constant-one function is $\mathbb{R}$, which is not compact by [F3] (equivalently, the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover). Hence $F^{\ast }\rho \mathrm{\notin }{\mathrm{\Omega }}_c^0(\mathbb{R})$. Pullback therefore fails even to define the proposed compact-support cochain map, so it cannot induce the claimed contravariant cohomology map.

Here $F$ is nonproper because the compact singleton $\{0\}$ has inverse image $\mathbb{R}$; this is exactly the obstruction isolated by [F1]. The zero input still pulls back to compact support and an empty source would be vacuous, but neither repairs the universal assertion. The example is already in degree zero and dimension one, has no boundary endpoints, and uses one explicit bump with no choice principle.