Degree is invariant under proper smooth homotopy

Each $F_t$ is proper: for compact $K\subset N$, the closed slice $H^{-1}(K)\cap (M\times \{t\})$ is compact and projects homeomorphically onto $F_t^{-1}(K)$. Let $\omega \in {\mathrm{\Omega }}_c^n(N)$. Since $d\omega =0$ in dimension $n$, [F3] gives $F_1^{\ast }\omega -F_0^{\ast }\omega =d(K{H}^{\ast }\omega )\mathrm{.}$ If $C=H^{-1}(\mathrm{supp}\omega )$, then $C$ is compact. The form $K{H}^{\ast }\omega$ vanishes outside the compact projection ${\mathrm{pr}}_M(C)$, because its defining time integral has zero integrand there. Thus the displayed primitive has compact support.

For $n\ge 1$, [F2] applied to this compactly supported primitive yields ${\int }_M{F}_1^{\ast }\omega ={\int }_M{F}_0^{\ast }\omega$. The defining identity [F1], applied to an integral-one top class, therefore gives $\mathrm{deg}(F_1)=\mathrm{deg}(F_0)$. If $n=0$, connected $M$ and $N$ are points, so $F_0=F_1$ and the equality is immediate. Empty manifolds are excluded, the two endpoints are both checked, and all compactness operations use the one supplied compact set $C$; no family is selected and no AC is used.