 Solution

For compact connected oriented $n$-manifolds without boundary, the degree is characterized byfor every smooth top-degree form $\eta$. Equivalently, at any regular value $y$,The preimage is finite by compactness and the inverse function theorem. Taking a form supported in a small neighbourhood of a regular value and applying change of variables proves the signed-count formula. The top de Rham cohomology of a compact connected oriented manifold is one-dimensional, with integration giving its coordinate; pullback and Stokes theorem make the integral characterization independent of the chosen normalized form. Thus this signed count is independent of the regular value. In particular a nonzero degree forces surjectivity, and the number of points in any regular fibre is at least the absolute degree. Degree is invariant under homotopy, as follows by applying Stokes theorem to the pullback on $M\times [0,1]$.

For the polynomial assertion take $k\ge 1$. The map $P:\mathbb{C}\to \mathbb{C}$ extends smoothly to a map of the Riemann sphere fixing infinity: in reciprocal coordinates its target coordinate isThe homotopy $P_t(z)=z^k+t(a_1{z}^{k-1}+\cdots +a_k)$ extends smoothly in the same way. The map $z\mapsto z^k$ has $k$ distinct inverse images of the regular value one, each orientation-preserving, since the real Jacobian of multiplication by a nonzero complex derivative is its squared absolute value. Its degree is $k$, so $\mathrm{deg}P=k\ne 0$. It is onto the sphere and must attain zero at a finite point. This proves every nonconstant complex polynomial has a root, without assuming the result to count roots of a general polynomial.

For a proper smooth $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$, choose a compactly supported $n$-form $\eta$ with $\int \eta =1$ and defineProperness makes its pullback compactly supported. If ${\eta }^{\prime }$ is another choice, the given result gives $\eta -{\eta }^{\prime }=d\beta$ with compact support. The difference of the defining integrals is $\int d(F^{\ast }\beta )=0$, proving independence. Sard's theorem supplies a regular value. A proper map has finitely many preimages of it, and a sufficiently small neighbourhood has no additional inverse sheets outside the chosen inverse-function neighbourhoods: otherwise a sequence of extra preimages would have a convergent subsequence by properness and limit in that fibre. A normalized form there therefore gives the same finite signed-count formula, proving that the degree is an integer.

Finally $x\mapsto x^k$ is proper for $k\ge 1$. For odd $k$, a positive regular value has one inverse image with positive derivative. For positive even $k$, it has two inverse images with opposite derivative signs. ConsequentlyFor $k=0$ it is the constant map one, whose inverse image of the compact singleton $\{1\}$ is all of $\mathbb{R}$. It is not proper, so its degree is not defined by the proper-map construction requested here.