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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.

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By the Hirsch-Smale h-principle, isotopy classes of immersions are determined by their Gauss maps. Since the target has trivial tangent bundle and the source stably trivial normal bundle, there is no obstruction, so there are immersions. – Ben Wieland May 19 '10 at 17:54
Andre, your third paragraph is a profoundly lazy and lame answer. I wish I'd thought of it! – Allen Knutson May 27 '10 at 1:25
After reading Ben's comment, I've tried to find the immersion, but without success. You can think of that immersion as an S^2-parametrized family of 1-manifolds, where singular fibers are allowed. – André Henriques May 27 '10 at 16:47

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