Suppose $k \in \mathbb{Z}$. I would like to find a smooth map $f : \mathbb{S}^n \to \mathbb{S}^n$ such that $\deg(f)=k$.
I tried with taking $k$ sets in $\mathbb{S}^n$ diffeomorphic to $\mathbb{D}^n$ and taking $f$ to be ugual to: $$g : \mathbb{D}^n \to \mathbb{S}^n \mbox{ with } g(x)=(2\sqrt{1-\|x\|^2}x, 2 \|x\|^2-1)$$
in these sets diffeomorphic to $\mathbb{D}^n$ but I can only prove that $f$ is continuos.
I thought that using approximation with smooth functions could help but I do not know how.
Here the degree of a map is defined as in Milnor's Topology from the differentiable viewpoint.
Here’s a recursive construction. Suppose you know $g: S^{n-1} \rightarrow S^{n-1}$ continuous of degree $k$. Consider $f:S^n \rightarrow S^n$ given by $f((\epsilon,0) \in \{\pm 1\} \times D^n) =(\epsilon,0)$, and $f((u,x) \in (-1,1) \times \sqrt{1-u^2}S^{n-1})=\left(u,\sqrt{1-u^2}g\left(\frac{x}{\sqrt{1-u^2}}\right)\right)$.
(Basically you apply $g$ to each “parallel” of $S^n$).
If $\ell$ is a smooth function on $S^n$, let $l_u(x \in S^{n-1}=l(u,sqrt{1-u^2}x)$. Then, for some (unimportant) weight function $w$: $$\int_{S^n}{\ell \circ f}=\int_{-1}^1{w(u)\int_{S^{n-1}}{(\ell\circ f)_u}\,du}.$$
Now, if $x \in S^{n-1}$ and $-1<u<1$, then $(\ell \circ f)_u(x)=\ell(f(u,\sqrt{1-u^2}x))=\ell(u,\sqrt{1-u^2}g(x))=\ell_u\circ g(x)$. So $$\int_{S^{n-1}}{(\ell \circ f)_u}=\int_{S^{n-1}}{\ell_u \circ g}=k\int_{S^{n-1}}{\ell_u}.$$
Finally, $$\int_{S^n}{\ell \circ f}=\int_{-1}^1{w(u)k\int_{S^{n-1}}{\ell_u}\,du}=k\int_{S^n}{\ell},$$ so $f$ has degree $k$. Of course, even if $g$ is smooth, $f$ might not be smooth – something may happen at the poles (the $(\epsilon,0)$, $\epsilon=\pm 1$). However, even if $g$ is not smooth, any smooth small enough perturbation of $f$ (and they exist) will still be of degree $k$.
To conclude, it’s enough to construct on $S^1$ a function $f$ of degree $k$, but such is $z \longmapsto z^k$.