Proving that hopf map from $S^3 \to S^2 $ is not null homotopic I want to prove that hopf map from $S^3 \to S^2 $ is not null homotopic. Is there some elementary proof of this fact?
 A: If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
On the other hand, the Hopf map is the attaching map of the $4$-cell in $\Bbb CP^2$, so its mapping cone is just $\Bbb CP^2$.
A: Briefly: This follows since the Hopf map $\pi:S^3\to S^2$ is surjective and satisfies the homotopy lifting property: if $\pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $\text{id}_{S^3}$ to a non-surjective map, which is impossible. 
In more detail: The Hopf map $\pi\colon S^3\to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $\pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $\pi$ satisfies the homotopy lifting property. 
Assume (to obtain a contradiction) that there is a nullhomotopy $\pi_t:S^3\to S^2$ with $\pi_0 = \pi$ and $\pi_{1}$ a constant map.
By the homotopy lifting property, there exists a homotopy $h_t\colon S^3 \to S^3$ satisfying $h_0 = \text{id}_{S^3}$ and $\pi\circ h_t = \pi_t$ for all $t$. Since $\pi$ is surjective and $\pi_1$ is not, $\pi \circ h_1 = \pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $\text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $\text{deg}(h_1) = \text{deg}(h_0) = \text{deg}(\text{id}_{S^3}) = 1$, so we have obtained a contradiction. 
A: The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.
