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 $E$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.