Nonzero degree forces surjectivity on closed oriented manifolds

Suppose $f$ omits a point $y\in N$. Represent $[M]$ by a finite singular cycle $c$. Every simplex of $f_{\mathrm{\#}}c$ lies in $N\setminus \{y\}$, so [F2] gives $q_y(f_{\ast }[M])=0$. But [F1] gives $f_{\ast }[M]=\mathrm{deg}(f)[N]$, hence $0=\mathrm{deg}(f)q_y([N])$. Since $q_y([N])$ is a generator of an infinite cyclic group by [F1], this forces $\mathrm{deg}(f)=0$, contrary to the hypothesis. Therefore $f$ is surjective. This uses the integral homological degree throughout and does not require a comparison with compact-support degree.

For the concrete map, [F3] gives $\mathrm{deg}(P_3)=3\mathrm{\ne }0$, so step 1.1 makes it surjective. The fibres can also be seen directly: for any $[a]\in \mathbb{R}\mathrm{/}\mathbb{Z}$, the three classes $[a\mathrm{/}3]$, $[(a+1)\mathrm{/}3]$, and $[(a+2)\mathrm{/}3]$ map to $[a]$; reducing $3t-a\in \mathbb{Z}$ modulo $3$ proves that these are all the preimages. Hence this example calculates rather than merely naming the surjectivity conclusion.

The nonzero hypothesis is essential: [F3] gives the constant map $P_0$ degree zero, and it omits every point other than $[0]$. In dimension zero, nonempty connected source and target are singletons and the unique map is already surjective, as [F1] records; the relative-chain contradiction in step 1.1 also applies there. Empty or disconnected manifolds lie outside the scalar degree hypotheses, and there are no manifold-boundary endpoints because closed manifolds are boundaryless. Only one omitted point and one representative cycle are used, so no choice axiom is used here.