Degree is well defined and independent of the normalized top form

Put $d={\int }_M{F}^{\ast }\nu$, which is defined because [F3] preserves its compact support. For another compactly supported top form $\omega$, let $a={\int }_N\omega$. By [F5], $\omega -a\nu$ has integral zero; by [F2] it equals $d\eta$ with $\eta$ compactly supported in degree $n-1$. Therefore [F4] gives $F^{\ast }\omega -a{F}^{\ast }\nu =d(F^{\ast }\eta )\mathrm{.}$ By [F3], the primitive on the right is compactly supported, so [F5] and linearity give ${\int }_M{F}^{\ast }\omega =ad$. This calculation uses compact support on $\eta$, not merely on its derivative. For $n=0$, [F2] instead says $\omega -a\nu =0$ and the same equation follows with the zero negative-degree primitive.

If ${\nu }^{\mathrm{\prime }}$ is any other normalized form, substitute $\omega ={\nu }^{\mathrm{\prime }}$ and $a=1$ in step 1.1 to obtain ${\int }_M{F}^{\ast }{\nu }^{\mathrm{\prime }}=d$. If a scalar $d^{\mathrm{\prime }}$ satisfies the asserted identity for every $\omega$, substitution of the originally chosen $\nu$ gives $d^{\mathrm{\prime }}={\int }_M{F}^{\ast }\nu =d$. Thus the identity defines a unique scalar independent of normalization. On the class $[\nu ]={\mathrm{Int}}_N^{-1}(1)$, the composite used in [F1] has exactly the value $d$, so it agrees with the already named degree. This argument derives existence and uniqueness from [F2]–[F5]; [F1] is used only to identify the notation.

Zero $\omega$ gives $a=0$ and no division by $a$; a degree-zero map presents no exception. At $n=0$, [F1] computes the two point orientation signs, and step 1.1 treats the zero negative-degree space directly. At $n=1$, $\eta$ is a compactly supported function, and [F3] is precisely what makes its pullback an admissible function primitive for compact Stokes. There are no boundary endpoints because both manifolds are boundaryless. Nonemptiness is needed for the normalized form. Only one such form and one primitive for the input difference were selected, not a family of them; [F2]–[F5] require no AC.