I know that $\mathbb{R} ^ 2$ minus a point deformation retracts onto a circle

What I'm looking to prove is that $\mathbb{R}^2$ minus $n$ points deformation retracts onto the wedge sum of $n$ circles, without explicitly stating the deformation retraction

I can see it by drawing it, but I'm not sure how to go about proving it. I thought about proving that since $\mathbb{R} ^ 3 $ minus parallel lines is homeomorphic to $\mathbb{R}^2$ minus $n$ points and since $\mathbb{R} ^ 3 $ minus parallel lines deformation retracts onto the wedge sum of $n$ circles then so does $\mathbb {R}^2$ minus $n$ points, but in this case I dont know how to prove that $\mathbb{R} ^ 3 $ minus parallel lines deformation retracts onto the wedge sum of $n$ circles

Any help would be much appreciated

  • 1
    $\begingroup$ To say you want to prove this "without explicitly stating the deformation retraction" is like saying you want to put that nail in the wood "without explicitly hitting it on the head with a hammer". $\endgroup$
    – Lee Mosher
    Apr 15 at 12:36
  • $\begingroup$ Okay so what would be the deformation retraction? $\endgroup$ Apr 15 at 12:45
  • $\begingroup$ $\mathbb{R} ^3$ minus $n$ parallel lines deformation retracts onto a sphere with 2$n$ points removed which then deformation retracts onto the wedge sum of 2$n$ -1 circles, am I getting somewhere with this? $\endgroup$ Apr 15 at 12:48
  • $\begingroup$ I don't think that going up a dimension to $\mathbb R^3$ is going to help you at all; it'll just muck things up. Also, one knows that open subsets of $\mathbb R^2$ are not homeomorphic to open subsets of $\mathbb R^3$ (by the Invariance of Domain Theorem of Algebraic Topology). $\endgroup$
    – Lee Mosher
    Apr 15 at 13:21
  • $\begingroup$ I would just try to deeply understand the construction of the deformation retraction for the case $n=1$ which you say you know, and then try to generalize that construction to arbitrary $n$. $\endgroup$
    – Lee Mosher
    Apr 15 at 13:23

When you say "the wedge", I interpret this as "a wedge" i.e. a subset of the plane homeomorphic to the wedge of circles.

Since this is likely to be a homework, I will only give you a sequence of steps:

  1. If $X, Y$ are any two finite subsets of the same cardinality in $R^2$, then there exists a homeomorphism $R^2\to R^2$ which carries $X$ to $Y$.

This, it does not matter which of the two finite subsets of the plane you have to consider. (This part of the problem was already solved at least once on MSE.)

  1. Suppose that $D$ is an open bounded convex subset of $R^2$. Then for every $x\in D$ there exists a deformation-retraction $D-\{x\}\to \partial D$.

  2. There exists a wedge of $n$ circles $$ W= \vee_{i=1}^n C_i $$ in $R^2$ such that each circle $C_i$ bounds a convex domain $D_i$ in $R^2$. Here, all the circles intersect at $0$. For each $i$ pick a point $x_i\in D_i$.

  3. Let $S^2= R^2\cup\{\infty\}$, the Riemann sphere (the 1-point compactification of $R^2$). Let $D_0$ denote the component of $S^2 - W$ containing the point $\infty$ and let $\bar{D}_0$ denote its closure. Show that there exists a continuous map of the closed disk $f: \bar{D}^2\to \bar{D}_0$ such that:

(a) $f: D^2\to D_0$ is a homeomorphism.

(b) $f$ is 1-1 overall except for $n$ boundary points of $D^2$ which are all mapped to $0\in \bar{D}_0$.

(In fact, you only need Part (a) but any reasonable construction will satisfy (b) as well.)

If you want to write a detailed proof, this is the hardest step.

  1. Use 4 to show the existence of a deformation retraction $D_0-\{\infty\}\to \partial D_0= W$. Use 2 to construct deformation retractions $D_i-\{x_i\}\to C_i$ for each $i=1,...,n$.

  2. Verify that all these deformation retraction taken together yield a deformation retraction $$ R^2 - W\to R^2- X, $$
    where $X=\{x_1,...,x_n\}$.

  • $\begingroup$ An alternative is this: For a closed circle segment $S \subsetneqq S^1$ let $C(S)$ be the convex hull of $S$ and $0$ and $R(S)$ be the convex hull of $\partial S$ and $0$ (which consist of two radial line segments from $0$ to a point of $S^1$). Show that $C(S)$ strongly deformation retracts to $R(S)$. Now take $n$ pairwise disjoint closed circle segments $S_i$ on $S^1$. Then $\mathbb R^2$ strongly deformation retracts to $D^2$ and $D^2$ strongly deformation retracts to $\bigcup_{i=1}^n C_i$ = wedge of the $C_i$ on $\mathbb R^2$. $\endgroup$
    – Paul Frost
    Apr 16 at 12:34

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.