# 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?

• What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined. – Justin Young Oct 9 '16 at 15:57
• @Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book) – happymath Oct 11 '16 at 4:50
• In that case, the answer below is best. The key point is that the cup product in $\mathbb CP^2$ is non-trivial, as opposed to $S^2 \vee S^4$. – Justin Young Oct 11 '16 at 14:05

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 $\Bbb CP^2$.
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.