(Explicitly) Constructing Deformation Retractions

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.