Degree is multiplicative under composition

The composite is proper because for compact $K\subset P$, first $G^{-1}(K)$ and then $F^{-1}(G^{-1}(K))$ are compact. For $\omega \in {\mathrm{\Omega }}_c^n(P)$, [F2] and [F1] give ${\int }_M(G\circ F{)}^{\ast }\omega ={\int }_M{F}^{\ast }(G^{\ast }\omega )=\mathrm{deg}(F){\int }_N{G}^{\ast }\omega =\mathrm{deg}(F)\mathrm{deg}(G){\int }_P\omega \mathrm{.}$ The uniqueness clause of [F1] proves the displayed composition formula, including when either factor is zero.

For the identity, ${\int }_M{\mathrm{id}}^{\ast }\omega ={\int }_M\omega$, so [F1] gives degree $1$. If all three manifolds are closed, [F3] identifies each scalar in step 1.1 with its homological degree, and the resulting equality is precisely the choice-free clause of [F4]; its separate AC-dependent top-cohomology clause is not used. In dimension zero the formula multiplies the source/intermediate and intermediate/target orientation signs, so the intermediate sign squares to $1$. Empty manifolds are excluded, degree-zero maps and identity endpoints were included above, and the proof makes no choice of forms because [F1] is an identity for every supplied form.