Every integer is realized by a map to the sphere

(The bump profile.) By [F1] in dimension one fix a smooth $\chi :\mathbb{R}\to [0,1]$ with $\chi =1$ on $[-1\mathrm{/}2,1\mathrm{/}2]$ and $\chi =0$ outside $(-3\mathrm{/}4,3\mathrm{/}4)$, and define $W(s):=1-\chi (s{)}^2(1-s)$ for $s\ge 0$. Then $W$ is smooth, $W(s)=s$ for $0\le s\le 1\mathrm{/}2$, $W(s)=1$ for $s\ge 3\mathrm{/}4$, and $0\le W(s)\le 1$ with $W(s)>0$ for $s>0$; moreover $1-W(s)=\chi (s{)}^2(1-s)$ is the square of the smooth function $\delta (s):=\chi (s)\sqrt{1-s}$, read as $0$ where $\chi$ vanishes.

(The model pinch.) Put $G(s):=W(s)\mathrm{/}s$ for $s>0$ and $G(0):=1$; since $W(s)=s$ for $s\le 1\mathrm{/}2$, the function $G$ is smooth and positive on $[0,\mathrm{\infty })$. Define $\mathrm{\Phi }(s):=\chi (s)\sqrt{W(s)(1-s)\mathrm{/}s}$ for $0<s\le 3\mathrm{/}4$ and $\mathrm{\Phi }(0):=1$ and $\mathrm{\Phi }(s):=0$ for $s\ge 3\mathrm{/}4$; the radicand $Q(s)=W(s)(1-s)\mathrm{/}s$ is smooth and positive on $[0,3\mathrm{/}4)$ and equals $1-s$ near $0$, so $\sqrt{Q}$ is smooth and positive there by [F2], and since $\chi$ is flat at $3\mathrm{/}4$ and vanishes beyond it, $\mathrm{\Phi }=\chi \sqrt{Q}$ is smooth on $[0,\mathrm{\infty })$. Set $F(x):=(2\mathrm{\Phi }(\parallel x{\parallel }^2)x,2W(\parallel x{\parallel }^2)-1)$ for $x\in {\mathbb{R}}^m$. With $s=\parallel x{\parallel }^2$ one has $\mathrm{\Phi }(s{)}^{2}s=\chi (s{)}^{2}W(s)(1-s)=W(s)(1-W(s))$, hence $\parallel F(x){\parallel }^2=4\mathrm{\Phi }(s{)}^{2}s+(2W(s)-1{)}^2=4W(1-W)+4W^2-4W+1=1$, so $F$ maps into $S^m$. Each component is smooth, and since $F$ is locally constant with value $N=(0,\dots ,0,1)$ for $s\ge 3\mathrm{/}4$, its expressions in the two stereographic charts of $S^m$ ([F3]) are smooth; thus $F:{\mathbb{R}}^m\to S^m$ is smooth.

(The regular value of the model.) At the origin $W(0)=0$ and $\mathrm{\Phi }(0)=1$, so $F(0)=(0,\dots ,0,-1)=:y_{-}$. If $F(x)=y_{-}$ then $2W(s)-1=-1$, that is $W(s)=0$, which by step 1.1 happens only for $s=0$; hence $F^{-1}(y_{-})=\{0\}$. Near $0$ one has $W(s)=s$, $G(s)=1$ and $\mathrm{\Phi }(s)=\chi (s)\sqrt{1-s}$, so $\mathrm{\Phi }(0)=1$; the first $m$ components of $F$ have differential $2\mathrm{\Phi }(0)\mathrm{i}\mathrm{d}=2\mathrm{i}\mathrm{d}$ at $0$ and the last component has vanishing differential there, so $d{F}_0$ has rank $m$ and is an isomorphism of tangent spaces, and $y_{-}$ is a regular value of $F$.

(Gluing the model into $M$.) For $k\mathrm{\ne }0$ put $n=\mathrm{\mid }k\mathrm{\mid }$. Since $M$ is nonempty and $m\ge 1$, choose one chart ball and $n$ distinct coordinate points in it. Choose sufficiently small pairwise disjoint open Euclidean balls about these points with closures still inside that chart ball. Translation and positive rescaling give charts ${\phi }_i:U_i\to B_2(0)$ with ${\phi }_i(p_i)=0$, pairwise disjoint domains $U_i$, and closed unit coordinate balls $B_i={\phi }_i^{-1}({\bar{B}}_1(0))\subset U_i$. Each $B_i$ is compact as the continuous image of a compact Euclidean ball, and hence closed in the Hausdorff $M$. Define $f=F\circ {\phi }_i$ on $U_i$ and $f=N$ on the open complement of ${\bigcup }_i{B}_i$. The only overlaps are $U_i\setminus B_i$, on which $F=N$ by step 2.1; thus the definitions agree. Their smooth local expressions paste to a smooth $f$. This selects finitely many chart data and ensures disjoint domains, not merely disjoint closed balls.

(Degree of the glued map.) By steps 3.1 and 3.2 the equation $f(x)=y_{-}$ holds exactly for $x=p_1,\dots ,p_n$, and $d{f}_{p_i}=d{F}_0\circ d{\phi }_i$ is an isomorphism, so $y_{-}$ is a regular value of $f$ with finite fibre $\{p_1,\dots ,p_n\}$, and $f$ is proper because $M$ is compact ([given]). By [F4], $\mathrm{deg}(f)={\sum }_{i=1}^n\mathrm{sgn}(d{f}_{p_i})=\mathrm{sgn}(d{F}_0){\sum }_{i=1}^n{\epsilon }_i$, where ${\epsilon }_i=+1$ if ${\phi }_i$ preserves the orientations of $M$ and ${\mathbb{R}}^m$ and ${\epsilon }_i=-1$ otherwise; each chart is orientation-preserving or orientation-reversing, and composing a chart with $x\mapsto (-x_1,x_2,\dots ,x_m)$ reverses its orientation while fixing its centre and unit ball. At $y_{-}=-e_{m+1}$ the ambient tuple $(y_{-},2e_1,\dots ,2e_m)$ has sign $(-1{)}^{m+1}$, so $\mathrm{sgn}(d{F}_0)=(-1{)}^{m+1}$ for the outward-normal-first orientation. Choosing all ${\epsilon }_i$ equal to $\mathrm{sgn}(k)\mathrm{sgn}(d{F}_0{)}^{-1}$ gives $\mathrm{deg}(f)=k$; no infinite selection is used.

(The case $k=0$ and conclusion.) For $k=0$ take the empty family, so $f$ is the constant map $N$, whose regular values are the points different from $N$ and whose fibre is then empty; hence $\mathrm{deg}(N)=0$ by [F4]. For every $k\in \mathbb{Z}$ the map constructed in steps 3.2 and 4.1 is therefore smooth with $\mathrm{deg}(f)=k$, using finitely many charts and finitely many choices of closed balls and chart orientations only.