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  "body": "\n## Statement\n\nLet $n \\ge 1$, let $D^n = \\set{x \\in \\R^n : \\norm{x} \\le 1}$ be the [[closed unit ball|0Q0H13G8ZB]], and let $S^{n-1} = \\set{x \\in \\R^n : \\norm{x} = 1}$ be its boundary [[sphere|T62369RFH6]].\nThere is no [[continuous map|0VGFHTM75Y]] $r \\colon D^n \\to S^{n-1}$ with $r(x) = x$ for all $x \\in S^{n-1}$.\n\n## Remarks\n\nSuppose such a retraction $r$ exists, and let $i \\colon S^{n-1} \\to D^n$ be the inclusion, so that $r \\circ i = \\id_{S^{n-1}}$.\nApply [[reduced homology|NJQMWVK3QK]] in degree $n - 1$: the identity of $\\tilde{H}_{n-1}(S^{n-1}; \\Z) \\cong \\Z$ factors through $\\tilde{H}_{n-1}(D^n; \\Z) = 0$, which is impossible.\nFor $n = 1$ the argument is simply that $D^1 = [-1, 1]$ is connected and $S^0 = \\set{-1, 1}$ is not.\nFor smooth retractions there is also a proof by Sard's theorem: the preimage of a regular value would be a compact $1$-manifold with only one boundary point.\n\nExample of use: the theorem gives the Brouwer fixed-point theorem.\nIf $f \\colon D^n \\to D^n$ had no fixed point, the ray from $f(x)$ through $x$ would meet $S^{n-1}$ in a point $r(x)$, and $r$ would be a retraction of $D^n$ onto $S^{n-1}$.\n\nThe ball cannot be replaced by any space: the cylinder $S^{n-1} \\times [0, 1]$ retracts onto its end $S^{n-1} \\times \\set{0}$ by $(x, t) \\mapsto (x, 0)$.\n\n## Explanations\n\n### L1 human\n\nYou cannot push a drum skin onto its rim continuously while holding the rim fixed.\nSomewhere the skin would have to tear.\n\n### L2 human\n\nThe no-retraction theorem is equivalent to the Brouwer fixed-point theorem, and each implies the other with a short argument.\nIn homotopy terms, it says that the identity map of $S^{n-1}$ is not null-homotopic, and so it is a statement about the degree of the identity.\nSimilar retraction arguments, with homology or cohomology, show that the boundary of a compact manifold is never a retract of the manifold.\n",
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