There exists surjective map $f:S^n \to S^n$ of degree zero.
By Hopf theorem, $f$ is homotopic to some constant map $c_{y}:S^n \to \{y\}$.
Let the homotopy be $H(x,t)$, $H(x,0) = f(x)$, $H(x,1) = c_y(x) = y$, so $H(-,t)$ gives a deformation retraction from $S^n$ to a point, but such deformation retraction shouldn't exist, so what's the problem here?
Best Answer
Examples of maps from $S^n$ to $S^n$ which are null-homotopic are just not deformation-retractions. Think of a sphere trying to wrap itself around a second sphere. Keep the first sphere on the surface of the second, but "unwrap it", and once it's unwrapped, collapse it to a point. I've formalized the idea below.
Write a point on the sphere as $(x_0, x_1,\cdots,x_n)$ where $\sum_{i=0}^n x_i^2 = 1$. Write such a point as $x = x_0 e_0 + v$ where $v = \sum_{i=1}^n x_i e_i$ and $\{e_0,\cdots,e_n\}$ is the standard basis of $\mathbb R^n$. We can re-write $x = (\cos \theta) x_0 + (\sin \theta) v_0$ where $v_0$ has norm $1$ and lies in the subspace spanned by $\{e_1,\cdots,e_n\}$; the parameter $\theta$ can take values from $0$ to $\pi$ while $v_0 \in S^{n-1}$ is arbitrary. This parametrization $\alpha$ of $S^n$ by $[0,\pi] \times S^{n-1}$ is bijective on the subset where $\theta \in ]0,\pi[$.
Consider the map $f : [0,\pi] \to [0,\pi]$ defined by $f(x) = 2x$ if $x \in [0,\pi/2]$ and $f(x) = 2x-\pi$ if $x \in [\pi/2, \pi]$. This map is homotopic to the map that sends everything to $0$ just by the homotopy $h(x,t) = tf(x)$.
Now consider $g = f \times \mathrm{id}_{S_n}$. By construction, this is homotopic to the map $(x \mapsto 0) \times \mathrm{id}_{S_n}$.
Note that $\alpha: [0,\pi] \times S^{n-1} \to S^n$ sends $\{0\} \times S^{n-1}$ to $\{e_0\}$ and $\{\pi\} \times S^{n-1}$ to $\{-e_0\}$. So there is indeed a well-defined null-homotopic map $\tilde g : S^n \to S^n$ defined by $\tilde g(\alpha(\theta, v_0)) = \alpha(f(\theta), v_0)$. (The reason why this is well-defined even though $\alpha$ is not bijective is because on the set of points where the fibers of $\alpha$ is not a singleton, namely $\{0\} \times S^{n-1}$ and $\{\pi\} \times S^{n-1}$, the above definition is independent of the choice of point in the fiber.) The homotopy is basically given by $\tilde h(\alpha(\theta, v_0), t) = \alpha(tf(\theta), v_0)$.
Hope that helps,