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I was wondering how to prove that $\mathbb R^n$$-0$ deformation retracts to a sphere $\mathbb S^n$$^-$$^1$?

And also, why is the punctured disk contractible, i.e. why is $\mathbb D^2$$-${$(1,0)$} contractible?

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    Hint : take a segment between $x$ and $x/|x|$ – 2017-02-12
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    The punctured disk is not contractible. (unless you're removing a point on the boundary, but almost no one would call that punctured). It is homotopy equivalent with a circle. – 2017-02-12
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    But yes, I meant on removing a point from the boundary, with coordinates $(1,0)$ – 2017-02-12
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    I need something to be clarified: is question talking about $S$ to the power of $n-1$, or is it talking about something else? – 2017-02-12
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    Yes, exactly @关一俊 – 2017-02-12
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    $S^{n-1}$ does not mean $S$ to the power of $n-1$, it is simply notation for the $(n-1)$-sphere. – 2017-11-11

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For proving that $X=\mathbb R^n-\{0\}$ deformation retracts to $S^{n-1}$, take $$r:\mathbb R^n-\{0\}\longrightarrow S^{n-1}, \ r(x)=\frac{x}{\lVert x \rVert}$$ which is a retract, and now you have to prove that $i\circ r\equiv Id_X$. To do this, take the homotopy $$H:X\times [0,1] \longrightarrow X, \quad H(x,t)=tx+(1-t)\frac{x}{\lVert x \rVert}$$ Now, note that $H(x,t)\neq0$ for all $x\in X$, $t\in[0,1]$ so it is well defined. It is clear that $H$ is the desired homotopy.

For the other question, if you remove an inner point from the disc, it would be homotope to $S^1$, using the same reasoning as above. If you remove a boundary point, then $D-\{p\}$ is convex, so it's contractible.

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    You just proved there is a retraction, not a deformation retract. – 2017-02-12
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    I left that to the OP, but I'll complete it anyway. – 2017-02-12
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Here's a deformation retraction of $\mathbb{R}^N \setminus \{0\}$ to $S^{n-1}$. Define $\phi_t:\mathbb{R}^n \setminus \{0\} \to S^{n-1}$ by $$\phi_t(x) = (1-t)x + t \frac{x}{\|x\|}$$ Then $\phi_0$ is the identity on $\mathbb{R}^n \setminus \{0\}$ and $\phi_1$ maps to $S^{n-1}$.

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    Oh, I see it. Thanks! How about proving a punctured disk contraction.. – 2017-02-12
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    I don't think the punctured disk is contractible. – 2017-02-12
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    But what I meant, is disk with removing a point from the boundary, with coordinates $(1,0)$... Sorry for misunderstanding. – 2017-02-12
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    I think you could just do a straight-line deformation retraction onto some point on the boundary on the "opposite" side of the point removed. – 2017-02-12