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I'm having trouble building the actual deformation retractions, although I understand the concepts behind them, homotopies, etc.

For example, when constructing a deformation retraction for $\mathbb{R}^n-\{0\}$ to $S^{n-1}$, I found that you could define the mapping $F(x,t) = (\frac{x_1}{t||x||+(1-t)},...,\frac{x_n}{t||x||+(1-t)})$.
However, I still don't see how one thought of that in the first place. I get the idea of turning the $x_n$'s into unit vectors, but I don't understand the intuition behind the $+(1-t)$, etc.


Anyone want to give some advice on how you approach constructing such a family of functions?
In terms of an actual problem, I'm trying to construct a def. retraction of $T_2-\{p\}$ onto a graph with 2 circles intersecting in a point (the longitude/meridian circles of the torus). I understand why this is possible, but my intuition fails to construct the actual def. retraction.

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You might like $F(x, t) = (1 - t)x + tx/|x|$ better. Not sure! –  Dylan Moreland Sep 12 '11 at 3:57
    
I really don't like the deformation retraction you've found. The one Brian proposes is more intuitive, thanks to his explanation about what the function $tv+(1-t)u$ "does". Though, in this spirit, I would have written his deformation retraction without further simplifications: $t\frac{x}{\vert x \vert} + (1-t)x$. –  a.r. Sep 12 '11 at 5:05
    
^^ Yes, I like that one much better. Because of its form, it's so much easier to visualize. –  Dustin Tran Sep 12 '11 at 5:07

3 Answers 3

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The easiest deformation retracts to think of come from straight line homotopies. Note that the retract of $\mathbb{R}^n \setminus 0$ to $S^{n-1}$ given by Dylan Moreland is of this variety.

In this vein we can also answer your question. $T_2\setminus\{p\}$ is given by the quotient of the unit square with the point $(1/2,1/2)$ removed obtained by identifying opposite sides of the boundary. Try to show that the straight line homotopy of the unit square minus a point to its boundary induces in the quotient a deformation retract of $T_2 \setminus \{p\}$ to the wedge of two circles.

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This is very helpful! I'm starting to understand the intuition behind building def. retracts now! (or maybe just ones with straight lines) –  Dustin Tran Sep 12 '11 at 5:11

It may be helpful to point out that the function $\displaystyle F(x,t)=\frac{1}{t\vert x\vert+(1-t)}x$ is by no means the only one that does the job; I don’t even find it the most obvious straight-line homotopy.

I’d start from the fact that the function $f(t)=tv+(1-t)u$ traces out the segment from $u$ to $v$ as $t$ runs from $0$ to $1$. This is a basic, very useful fact that holds not just in $\mathbb{R}$, but in any real vector space. Thus, if $\alpha$ is any positive real, $$\varphi(t)=t\left(\frac{1}{\alpha}\right)+(1-t)\cdot 1=\frac{t}{\alpha}+1-t=\frac{t+(1-t)\alpha}{\alpha}$$ runs from $1$ to $1/\alpha$ as $t$ runs from $0$ to $1$, and $$F(x,t)=\left(\frac{t+(1-t)\vert x\vert}{\vert x\vert}\right)x$$ runs along the segment from $x$ to $\displaystyle\frac{x}{\vert x\vert} \in S^{n-1}$.

For this reason I’d focus more on the idea underlying the homotopy that you gave than on the details, beyond verifying that they actually work out; jspecter has already suggested how they might be applied to the torus question.

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Yes, I think it's a rare student who writes down the retraction from the punctured cell to the boundary square, although maybe that's a bad thing. –  Dylan Moreland Sep 12 '11 at 4:46
    
Thanks for the suggestion. I wasn't aware of that 'function fact' and it could be very useful to know. –  Dustin Tran Sep 12 '11 at 5:03
    
Are all real vector spaces convex, so that all points on the line are in the space? –  gary Sep 12 '11 at 5:42
    
@gary: If $V$ is a real vector space, and $u,v\in V$, every real linear combination of $u$ and $v$ is in $V$. What do you mean by convex space? Are you thinking of locally convex normed spaces? –  Brian M. Scott Sep 12 '11 at 7:44
    
@Brian: yes, you're right, I posted nonsense, I was thinking of convexity as metric/normed spaces. –  gary Sep 12 '11 at 9:20

You want a homotopy between the identity map on $R^n - \{0\}$ to the identity map on $S^{n-1}$ such that points on the sphere are fixed.

So essentially $F(x,t) = \frac{1}{t|x| + (1 - t)} x$. $F(x,0)$ is the identity map on $\mathbb{R}^{n} - \{0\}$ since $\frac{1}{0|x| + (1 - 0)} = 1$. On the other hand, $F(x,0)$ is on the sphere since $\frac{1}{1|x| + (1 - 1)} = \frac{1}{|x|}$, and $\frac{x}{|x|}$ has norm $1$. Clearly $F$ is continuous. Hence this the desired homotopy. Note that if $x$ is already on the sphere then $\frac{1}{t|x| + (1 - t)} = \frac{1}{t + (1 - t)} = 1$. Hence $F(x,t) = x$ for all $t$ between $0$ and $1$. Thus this homotopy fixes the points on $S^{n - 1}$.

Intuitively, the point $x$ is moving along the line between the origin and $x$ toward the point (i.e. $\frac{x}{|x|}$) on that line which is on the sphere.

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I understand why the function works. My question was how to (intuitively) come up with F to begin with. –  Dustin Tran Sep 12 '11 at 5:01
    
The idea is that you need to get every point in $\mathbb{R}^n - \{0\}$ to the circle in a continuous way. You know that $\frac{x}{|x|}$ is on the circle, but you can move it there too fast or it won't be continuous. So instead of dividing it by the norm right away, you slowly make the denominator closer to $|x|$. In some sense, $t|x| - (1 - t)$ slowly goes from $1$ to $|x|$. –  William Sep 12 '11 at 5:16

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