# Deformation Retraction and Projection/Closest Vector

How do I compute the projection/closest vector to a subset? I have been thinking about this for far too long without any progress. If it helps, I am working in $$\Bbb{R}^2$$, but I would like formalue in terms of norms and inner products, if possible.

For context, I am trying to prove that the figure eight is a deformation retract of the doubly punctured plane. And it is annoying that everything hinges on this annoyingly simple question. I have already shown that $$\overline{B}(0,1) \setminus \{p,q\}$$ is a deformation retraction of $$\Bbb{R}^2 \setminus \{p,q\}$$, where $$p = (-\frac{1}{2},0)$$ and $$q = (\frac{1}{2},0)$$. Now I just need to show that $$\overline{B}(0,1) \setminus \{p,q\}$$ deformation retracts to the union of the two discs with one centered at $$p$$, the other centered at $$q$$, but I am currently facing the obstacle discussed above.

EDIT: It just occurred to me that deformation retraction I had in mind won't be well-defined. Any point on the y-axis contained in $$\overline{B}(0,1)$$ won't have a unique projection/closest vector in the union of the two open discs contained in $$\overline{B}(0,1)$$...Hmm..need to rethink my approach...Of course, I wouldn't be opposed to any suggestions!

I do not really understand what you mean by "the projection/closest vector to a subset".

However, I shall explain how to get the desired strong deformation retraction. Let $$p_{\pm1}$$ denote the points $$(\pm1,0) \in \mathbb{R}^2$$, let the figure eight be the space $$E = S_{+1} \cup S_{-1}$$, where $$S_{\pm1}$$ is the circle around $$p_{\pm1}$$ with radius $$1$$ and let the doubly punctured plane be the space $$P = \mathbb{R}^2 \setminus \{ p_{+1}, p_{-1} \}$$. Define $$r : P \to E$$ as follows. For $$p = (x,y)$$ set $$r(p) = \begin{cases} p_{+1} + \dfrac{p - p_{+1}}{\lVert p - p_{+1} \rVert} & \lVert p - p_{+1} \rVert \le 1 \\ p_{-1} + \dfrac{p - p_{-1}}{\lVert p - p_{-1} \rVert} & \lVert p - p_{-1} \rVert \le 1 \\ \dfrac{2xp}{\lVert p \rVert^2} & \lVert p - p_{+1} \rVert \ge 1, p \ne 0, x \ge 0 \\ -\dfrac{2xp}{\lVert p \rVert^2} & \lVert p - p_{-1} \rVert \ge 1, p \ne 0, x \le 0 \end{cases}$$ Here $$\lVert - \rVert$$ denotes the Euclidean norm $$\lVert (x,y) \rVert = \sqrt{x^2+ y^2}$$. Note that the denominator $$\lVert p - p_{\pm 1} \rVert$$ does not vanish on $$P$$.

What happen geometrically? Let $$D_{\pm 1}$$ = closed unit disk with center $$p_{\pm 1}$$, $$H_{\pm 1}$$ = right/left half plane minus the interior of $$D_{\pm 1}$$.

The first two lines describe the radial strong deformation retractions from $$B_{\pm 1} = D_{\pm 1} \setminus \{ p_{\pm 1} \}$$ to $$S_{\pm 1}$$. In fact, for $$p \in B_{\pm 1}$$ we have $$\lVert r(p) - p_{\pm 1} \rVert = \lVert \dfrac{p - p_{\pm 1}}{\lVert p - p_{\pm 1} \rVert} \rVert = 1$$, and for $$p \in S_{\pm 1}$$ we have $$p_{\pm 1} + \dfrac{p - p_{\pm 1}}{\lVert p - p_{\pm 1} \rVert} = p$$.

Note that $$B_{+1} \cap B_{-1} = \{ 0 \}$$. Both line 1 and line 2 yield $$r(0) = 0$$.

The last two lines (together with $$r(0) = 0$$) describe strong deformation retractions of $$H_{\pm 1}$$ to $$S_{\pm 1}$$. This is done by shifting each point $$p \ne 0$$ along the line through $$0$$ and $$p$$ until it reaches $$S_{\pm 1}$$. To be formal, this line is given by $$l_p(t) = t p$$, and for $$p \in H_{\pm 1} \setminus \{ 0 \}$$ we must find $$t$$ such that $$\lVert t p - p_{\pm 1} \rVert = 1$$. Easy computations show $$t = \pm \dfrac{2x}{\lVert p \rVert^2}$$, and in fact we defined $$r(p) = l_p(t) = t p$$.

Note that for $$p \in Y = H_{+1} \cap H_{-1}$$ = $$y$$-axis = set of points with $$x = 0$$ we have $$r(p) = 0$$.

Thus all four lines give us a consistent definition on the whole space $$P$$. It remains to show that $$r \mid_{H_{\pm 1}}$$ is continuous in $$p = 0$$. We have $$\lVert p - p_{\pm 1} \rVert \ge 1$$, i.e. $$(x \mp 1)^2 + y^2 \ge 1$$. This is equivalent to $$\pm 2x \le x^2 + y^2$$ which means $$2\lvert x \rvert \le \lVert p \rVert^2$$ since $$p \in H_{\pm 1}$$. Hence $$\lVert r(p) \rVert = \dfrac{2 \lvert x \rvert}{\lVert p \rVert} \le \lVert p \rVert$$ for $$p \ne 0$$. This immediately implies continuity.

We now have constructed a retraction $$r$$. To see that it is a strong deformation retraction, define a homotopy $$H : P \times I \to P, H(p,t) = (1-t)p + tr(p) .$$ It is readily verified that in fact $$H(p,t) \ne p_{\pm 1}$$ for all $$(p,t)$$ (check all 4 lines in the definition of $$r$$). This is a homotopy from $$id_P$$ to $$r$$ which is stationary on $$E$$.