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Just getting started with Hatcher, wondering if I get the right idea at the beginning?

Problem 0.2, Page 18:

Construct an explicit deformation retraction of $\mathbb{R}^n - \{0\}$ onto $S^{n−1}$.

Consider $$f_t: X \to X, t \in I, f_t(x) = (1-t)x + t \frac{x}{|x|}.$$

Hence, $f_0 = \mathbb{I}$, $f_1 = \frac{x}{|x|}$, and $f_t|_{S^{n-1}} = S^{n-1}$ for all $t$.

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    $\begingroup$ Well, does it work? $\endgroup$ – Mariano Suárez-Álvarez Aug 30 '13 at 3:53
  • $\begingroup$ Yes I find it work, Dear @MarianoSuárez-Alvarez.... $\endgroup$ – 1LiterTears Aug 30 '13 at 4:02
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    $\begingroup$ Trying to figure out if @MarianoSuárez-Alvarez still wants to talk to me...? $\endgroup$ – 1LiterTears Aug 30 '13 at 4:08
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    $\begingroup$ @Jellyfish Strikes me as odd. $\endgroup$ – Pedro Tamaroff Aug 30 '13 at 4:17
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    $\begingroup$ @PeterTamaroff That's what I guessed too. It seems to me that it should be written either $f_t = t\mathbb{I} + (1-t)(x\mapsto x/|x|)$ or $f_t(x) = tx + (1-t)(x/|x|)$. $\endgroup$ – Neal Aug 30 '13 at 4:44
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Yes your solution is correct. What you could do additionally is to argue why $f\mid_{S^{n-1}}=id$ for all $t$. It can be argued as follows: if $x \in S^{n-1}$ then $x = {x \over |x|}$ and therefore $f(x) = (1-t)x + tx = x$.

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