Degree of a proper smooth map by compact-support cohomology

By [F1], ${\mathrm{Int}}_N$ is a bijective linear map, so its inverse is the function taking each real number to its unique preimage class. This uses uniqueness, not a choice of form representatives. The inverse is linear: applying the injective ${\mathrm{Int}}_N$ to the inverse image of $as+bt$ and to $a{\mathrm{Int}}_N^{-1}(s)+b{\mathrm{Int}}_N^{-1}(t)$ gives the same scalar. Together with [F2], the displayed composite is therefore a well-defined linear map $L:\mathbb{R}\to \mathbb{R}$.

Put $d=L(1)$. Every $t\in \mathbb{R}$ equals $t\cdot 1$, so linearity gives $L(t)=td$. For a compactly supported top form $\omega$, let $t={\mathrm{Int}}_N[\omega ]$. Then ${\mathrm{Int}}_N^{-1}(t)=[\omega ]$ and [F2] gives ${\int }_M{F}^{\ast }\omega =L(t)=d{\int }_N\omega \mathrm{.}$ Conversely, a scalar satisfying this equation for every such form equals $L(1)$ on any integral-one representative supplied by [F1]. Thus the composite definition and the unique-scalar characterization agree.

At $n=0$ every coordinate chart has singleton image in ${\mathbb{R}}^0$, so each point is open. A nonempty connected zero-manifold is therefore a single point, and $F$ is the unique map between the two points. If their orientation signs are ${\epsilon }_M,{\epsilon }_N$, then $F^{\ast }$ preserves the scalar value and $L(t)={\epsilon }_M{\epsilon }_{N}t$. This is consistent even when their signs differ. At $n=1$ [F2] preserves compact function primitives, as needed in the quotient. Zero forms give $0=d0$ and do not alone determine $d$; the integral-one class does. Empty manifolds are excluded because the target integration inverse would fail. Properness is exactly the support condition needed by [F2]; no regular value, compactness of $M$ or $N$, or choice axiom was assumed.