# Hopf fibration inside the retraction of R^4 minus line -> S^2?

This was inspired by this question.

Let $Y = {\mathbb R}^4 \setminus$a coordinate line, which retracts to ${\mathbb R}^3 \setminus$a point, which retracts to $S^2$.

What is an explicit immersion $S^3 \to Y$, whose composition with the above retraction gives the Hopf fibration?

My idea being, perhaps this would make clearer in what sense the $S^3$ is surrounding "a hole" in $S^2$.

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I can't tell if you can, but the most reasonable approach is that the Hopf fibration factors through the unit tangent bundle of the two sphere. The two sphere embeds in Y, in fact it embeds in $\mathbb{R}^3-\vec{0}$. That means the tangent space embeds, $(\mathbb{R}^3-\vec{0})\times \mathbb{R}^3$. Now try to project into $\mathbb{R}^3-\vec{0})\times \mathbb{R}=Y$ so that there is never an element of the tangent space of the unit tangent bundle in its kernel. However, it doesn't seem generic, so you need to "see" it to find the map. –  Charlie Frohman May 15 '10 at 13:50

First, note that $\mathbb R^4 \setminus \mathbb R \simeq S^2\times \mathbb R^2$. So you are asking for an immersion $S^3\to S^2\times \mathbb R^2$ representing the generator $\eta$ of $\pi_3(S^2\times \mathbb R^2)=\pi_3(S^2)=\mathbb Z$.
I'm guessing that your immersion doens't exist, and that you need to consider maps $S^3\to S^2\times \mathbb R^n$ with larger $n$, in order to represent $\eta$ by an immersion.
But if you are willing to go a little bit up in dimension, and consider maps $S^3\to S^2\times \mathbb R^4$, then you can even find an embedding representing $\eta$. It is given by $(H,I):S^3\to S^2\times \mathbb R^4$, where $H:S^3\to S^2$ is the Hopf map, and $I: S^3 \rightarrow {\mathbb R}^4$ is the standard inclusion.