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To determine this, recall that a continuous map $f: S^3 \to S^2$ is nullhomotopic if and only if its homotopy class $[f]$ is the zero element in the third homotopy group $\pi_3(S^2)$. The Hopf map is in fact a generator of $\pi_3(S^2) \cong \mathbb{Z}$, so $[f] \neq 0$.
One way to arrive at this conclusion is to view the Hopf map as the projection $p: S^3 \to S^2$ in the Hopf fibration $S^1 \hookrightarrow S^3 \xrightarrow{p} S^2$. This is a fiber bundle with fiber $S^1$. The long exact sequence in homotopy groups for this fibration is
\[
\cdots \to \pi_n(S^1) \to \pi_n(S^3) \to \pi_n(S^2) \to \pi_{n-1}(S^1) \to \cdots.
\]
For $n=3$, this gives
\[
\pi_3(S^1) \to \pi_3(S^3) \to \pi_3(S^2) \to \pi_2(S^1).
\]
It is a standard fact that $\pi_3(S^1) = 0$ (since $\pi_k(S^1) = 0$ for all $k > 1$) and $\pi_2(S^1) = 0$ (since $\pi_2(S^1)$ classifies maps from $S^2$ to $S^1$, which are nullhomotopic by extending over the ball). Another standard fact is that $\pi_3(S^3) \cong \mathbb{Z}$, generated by the homotopy class of the identity map $\mathrm{id}: S^3 \to S^3$. Substituting these values yields the short exact sequence
\[
0 \to \mathbb{Z} \to \pi_3(S^2) \to 0,
\]
so $\pi_3(S^2) \cong \mathbb{Z}$. Moreover, the middle map in the sequence is $p_*: \pi_3(S^3) \to \pi_3(S^2)$, the map on homotopy groups induced by $p$. Thus, $p_*$ is an isomorphism.
Now, $p_*([\mathrm{id}]) = [p \circ \mathrm{id}] = [p]$, the homotopy class of the Hopf map itself. Since $p_*$ is an isomorphism and $[\mathrm{id}]$ generates $\pi_3(S^3) \cong \mathbb{Z}$, it follows that $[p]$ generates $\pi_3(S^2) \cong \mathbb{Z}$. In particular, $[p] \neq 0$, so the Hopf map is not nullhomotopic.
An alternative (though related) approach is to use the Hopf invariant $H(f)$ of a map $f: S^3 \to S^2$, which is a homotopy invariant taking values in $\mathbb{Z}$. It may be defined cohomologically: if $f$ is smooth, suspend $f$ to a map $\Sigma f: S^4 \to S^3$ and let $X = S^3 \cup_{\Sigma f} e^5$ be the CW complex obtained by attaching a 5-cell via $\Sigma f$; then $H^*(X; \mathbb{Z}) \cong \mathbb{Z}[x,y]/(x^2, xy, y^2)$ with $|x|=3$ and $|y|=5$, and $H(f)$ is defined so that $\mathrm{Sq}^2(x) = H(f) y$ (where $\mathrm{Sq}^2$ is the Steenrod square). Constant maps have Hopf invariant 0. For the Hopf map itself, a direct computation shows $H(f) = \pm 1$ (depending on orientation conventions), so again $[f] \neq 0$.
A more geometric interpretation of the Hopf invariant is the linking number $\ell$ of the preimages $f^{-1}(a)$ and $f^{-1}(b)$ in $S^3$ for distinct regular values $a, b \in S^2$, with $H(f) = \pm \ell$ (again up to orientation). For the Hopf map, take $a = 0$ and $b = \infty$: then $f^{-1}(0) = \{(0,w) \in S^3 : |w|=1\}$ and $f^{-1}(\infty) = \{(z,0) \in S^3 : |z|=1\}$, which are great circles in $S^3$. These form the standard Hopf link, whose linking number is $\pm 1$ (computed, e.g., via the integral formula for linking number or by stereographic projection of $S^3$ to $\mathbb{R}^3$, where one circle maps to the unit circle in the $xy$-plane and the other to the $z$-axis union $\{\infty\}$). Thus, $H(f) = \pm 1 \neq 0$, confirming that the Hopf map is not nullhomotopic. |
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