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  "body": "\n## Example\n\nLet $S^3 = \\set{(z_0, z_1) \\in \\C^2 : \\abs{z_0}^2 + \\abs{z_1}^2 = 1}$, and identify $\\R^3$ with $\\C \\times \\R$.\nDefine $h \\colon S^3 \\to \\R^3$ by\n$$h(z_0, z_1) = \\bigl(2 z_0 \\overline{z_1}, \\ \\abs{z_0}^2 - \\abs{z_1}^2\\bigr).$$\nThe image lies in the unit sphere $S^2$, because\n$$\\abs{2 z_0 \\overline{z_1}}^2 + \\bigl(\\abs{z_0}^2 - \\abs{z_1}^2\\bigr)^2 = \\bigl(\\abs{z_0}^2 + \\abs{z_1}^2\\bigr)^2 = 1.$$\nTwo points of $S^3$ have the same image if and only if one is $(\\lambda z_0, \\lambda z_1)$ for the other $(z_0, z_1)$ and some $\\lambda \\in \\C$ with $\\abs{\\lambda} = 1$.\nSo each fibre of $h \\colon S^3 \\to S^2$ is a circle.\nThe map $h$ is a fibre bundle with fibre $S^1$, hence a [[Serre fibration|PB9JXQ2KC3]].\nEquivalently, $h$ is the quotient map $S^3 \\to \\C P^1 \\cong S^2$, $(z_0, z_1) \\mapsto [z_0 : z_1]$.\n\n## Remarks\n\nThe [[long exact sequence of homotopy groups|E0RV5D1GQ0]] of $S^1 \\to S^3 \\to S^2$ gives $\\pi_2(S^2) \\cong \\Z$ and $\\pi_3(S^2) \\cong \\pi_3(S^3) \\cong \\Z$.\nThe class of $h$ generates $\\pi_3(S^2)$, and $h$ has [[Hopf invariant|TEGTSY7XMX]] $\\pm 1$.\nAny two fibres are linked circles in $S^3$, with linking number $\\pm 1$.\nSimilar fibrations $S^3 \\to S^7 \\to S^4$ and $S^7 \\to S^{15} \\to S^8$ come from the quaternions and the octonions.\n\n## Explanations\n\n### L1 human\n\nThe Hopf fibration splits the $3$-sphere into circles, one circle over each point of the ordinary sphere.\nNo two of these circles are separate: every pair is linked like two rings of a chain.\nIt was the first example of a map from a sphere to a lower-dimensional sphere that cannot be deformed to a constant map.\n",
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