TITLE: (Explicitly) Constructing Deformation Retractions
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: 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.