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Reading Elon Lages Lima's book, during the proof of one theorem, he said that

So, given $k \in \mathbb{N}$, we may find $f_k \in \mathcal{C}^\infty (X,\mathbb{S}^n)$ such that $deg(f_k)=k$,

where, $X^n$ is a compact and orientable manifold, with $\partial X = \emptyset$. No other assumptions were made.

I mean, why so? It seems a very strong claim and I did not find (or was able to) provide a proof.

Anyway142
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  • For this to hold, it is sufficient to find the map $f_1$ and maps $g_k \in \mathcal C^\infty (\mathbb S^n,\mathbb S^n)$ with $\deg g_k = k$. The latter is easy to construct explicitly. I don't really know about the former. – SolubleFish Nov 28 '22 at 21:31
  • @SolubleFish do you mind providing more details about that? – Anyway142 Nov 28 '22 at 21:33
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    Given $f_1$, $g_k$ is easy, so $g_kf_1$ is what you want – FShrike Nov 28 '22 at 21:33
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    See this answer for constructing $g_k$: https://math.stackexchange.com/a/2312535/245133 – Sam Freedman Nov 28 '22 at 21:34
  • The above link only addresses continuous maps, but any continuous map is homotopic to a smooth map. – Didier Nov 28 '22 at 22:14

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Addressing $k\neq 0$, the case $k=0$ being trivial. First, if $X=S^n$, then there exist maps of any degree. Take, for instance, a map that rotates $k$ times around the equator.

Now, let $X$ be any compact oriented manifold of dimension $n$, without boundary. Take a map $h_k\colon S^n \to S^n$ of degree $k$, that is constant in a little neighbourhood $U$ of some point $p$. Quoting @SolubleFish: "There is a map $f\colon S^n\to S^n$ of degree one constant on some small neighborhood $U$ of a point $p$ (just squish $U$ onto $p$). Then if $g_k$ has degree $k$, $h_k=g_k\circ f$ has degree $k$ and is constant on a neighborhood of $p$."

Take the connected sum $X\#S^n$ with gluing region $D^n\subset U$, and consider the map $f_k$ which is constant equal to $h_k(p)$ on $X\setminus D^n$ and equal to $h_k$ on $S^n\setminus D^n$. This is a map of degree $k$. The result follows from the fact that $X\#S^n \simeq X$.

Didier
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  • Didier, how can you guarantee the existance of such $f_k$? – Anyway142 Nov 28 '22 at 21:58
  • There is a map $f:\mathbb S^n \to \mathbb S^n$ of degree one constant on some small neighborhood $U$ of a point $p$ (just squish $U$ onto $p$). Then if $g_k$ has degree $k$, $g_k\circ f$ has degree $k$ and is constant on a neighborhood of $p$ – SolubleFish Nov 28 '22 at 22:02
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    Here's an alternate approach. For, note that there is a map $f:X\rightarrow S^n$ of degree $\pm 1$: take any point $p$ with Euclidean neighborhood $U$, and map $U$ onto $S^{n}\setminus{pt}$ using stereographic projection, while mapping all points outside of $U$ to ${pt}$. Apart from the preimage of ${pt}$, every other preimage consists of a single point, which implies the degree is $\pm 1$. Now, note that $S^n$ admits self maps $g_k$ of any degree $k\in \mathbb{Z}$. One can simply suspect the $k$-power map $S^1\rightarrow S^1$ a bunch of times. Finally, use the maps to get all degrees. – Jason DeVito - on hiatus Nov 28 '22 at 22:15
  • @JasonDeVito This is much more elegant than my handcrafted proof! – Didier Nov 28 '22 at 22:26
  • That's a good point - @Didier's gives explicit smooth representatives. The stereographic trick of my comment can be smoothed to work, but that does require a bit of effort. (And I meant "suspend" instead of "suspect" in the penultimate sentence, and I ran out of space, but the last line should read "use the maps $g_k \circ f$" to get all degrees.) – Jason DeVito - on hiatus Nov 28 '22 at 22:45