A nonzero-degree map to a connected manifold is surjective

First $F[M]$ is closed. If $y\mathrm{\notin }F[M]$, [F2] gives an open neighbourhood $V$ of $y$ whose closure $K$ is compact. Properness makes $C=F^{-1}(K)$ compact, so [F3] makes $F[C]$ compact and closed in $N$. Since $F[M]\cap V=F[C]\cap V$, the open set $V\setminus F[C]$ contains $y$ and misses $F[M]$. Thus every point of the complement has an open neighbourhood in the complement.

Suppose $n\ge 1$ and $F$ is not surjective. Fix $y\in N\setminus F[M]$. By [F2] inside the open complement from step 1.1, choose an oriented coordinate ball $U$ containing $y$ whose closure is compact. By [F4] there is a smooth $\rho :N\to [0,1]$ with $\rho (y)=1$ and support contained in $U$. The support is closed by definition and lies in the compact set $\bar{U}$, hence is compact. In the positive chart $\varphi :U\to {\mathbb{R}}^n$, define the global top form $\omega$ by $\omega =\rho {\varphi }^{\ast }(d{x}^1\wedge \cdots \wedge d{x}^n)$ on $U$ and by zero off $U$; containment of the support in $U$ makes the two formulas agree smoothly near the edge of $U$.

Continuity and $\rho (y)=1$ give a nondegenerate closed coordinate rectangle $Q$ about $\varphi (y)$ on which $\rho \circ {\varphi }^{-1}\ge 1\mathrm{/}2$. On a larger bounding rectangle for the compact coordinate support, [F6] and the defining rectangular sum for the constant function give $I_{\varphi }(\omega )\ge \frac{1}{2}\mathrm{vol}(Q)>0.$ By [F5], $a:={\int }_N\omega =I_{\varphi }(\omega )>0$. Hence $\nu =a^{-1}\omega$ is compactly supported and has integral one.

The support of $\nu$ lies in $U\subseteq N\setminus F[M]$, so $F^{\ast }\nu =0$. Applying [F1] gives $0={\int }_M{F}^{\ast }\nu =\mathrm{deg}(F){\int }_N\nu =\mathrm{deg}(F)$, contrary to the hypothesis. Thus $F$ is surjective for $n\ge 1$. If $n=0$, connected nonempty $M$ and $N$ are singletons, and their unique map is already surjective. Empty manifolds are excluded; the zero-degree case makes no assertion. Only one missed point, one chart and one bump are selected, so no choice axiom is used.