I am attempting to find an explicit deformation retract from the torus minus some point to a figure of eight. Thus far, I have realised it is sufficient to show that:

If $I=[-1,1]$, then $I^2 - \{0,0\}$ deformation retracts to $\partial I^2$, since if we identify the edges of the square, we have the torus and the boundary of the square represents two circles intersecting at a single point, which is homeomorphic to a figure of 8.

I have also got the deformation retraction: $f_t(x,y)= \begin{cases} (x+t(1-x),y) &\quad\text{if $|x| > |y|$} \\ (x,y+t(1-y)) &\quad\text{if $|x| < |y|$} \\ \end{cases}$

I am not sure what to do if $|x|=|y|$. (Sorry if it is obvious!)

  • 1
    $\begingroup$ For any point $(x,y)$ in your punctured square, there is exactly one line going through $(0,0)$ and $(x,y)$. Considering sliding along this line, away from $(0,0)$. $\endgroup$ – Steve D Nov 11 '17 at 18:50

Here's an explicit formula, based on my comment above.

Let's suppose $|y|>|x|$. Then we want to linearly slide the $y$-coordinate from $y$ at $t=0$ to $sgn(y)$ at $t=1$. That's easy: $t\frac{y}{|y|}+(1-t)y$. According to my comment, we also want to scale $x$ by the same factor, to preserve the slope of the line from $(0,0)$ to $(x,y)$. So we get $t\frac{x}{|y|}+(1-t)x$. Of course, this works for $|x|>|y|$ very similarly, and so our homotopy is $$ f_t(x,y)=\begin{cases}(t\frac{x}{|y|}+(1-t)x, t\frac{y}{|y|}+(1-t)y)& |y|>|x|\\ (t\frac{x}{|x|}+(1-t)x, t\frac{y}{|x|}+(1-t)y)& |x|>|y| \end{cases} $$ And it's clear these branches agree when $|x|=|y|$.

EDIT: You can write the above deformation without branching:

$$ f_t(x,y) = \left(\frac{t}{\max(|x|,|y|)}+1-t\right)(x,y) $$

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    $\begingroup$ Would it be neater perhaps to write it in terms of $max\{|x|,|y|\}$? I know it doesn't really matter, but it looks a little nicer. $\endgroup$ – J.Jones5552 Nov 11 '17 at 19:39
  • $\begingroup$ @J.Jones5552: Good idea, done. $\endgroup$ – Steve D Nov 11 '17 at 21:24

I'm not sure your deformation retraction works, since what it appears to be doing is drawing a $\times$-shape in the square, moving the points in the top and bottom triangles vertically and moving the points in the left and right triangles horizontally. This means that the $\times$-shape itself gets (discontinuously) 'ripped', since its points want to move both vertically and horizontally.

I'd suggest a deformation retraction which moves points radially. Specifically, each $p \in I - \{ 0,0 \}$ there is a unique $\theta \in [0,2\pi)$ and $r>0$ such that $p = (r\cos \theta, r\sin \theta)$. Your deformation retraction could move $p$ along the ray $\{ (r\cos\theta, r\sin\theta) \mid r>0 \}$ until it hits $\partial I$, at which point it stops.

  • $\begingroup$ So rather than making a square hole, you make a circular one which terminates when you get to the edge? $\endgroup$ – J.Jones5552 Nov 11 '17 at 18:56
  • $\begingroup$ Right. Writing it down is a bit fiddly, but not impossible. $\endgroup$ – Clive Newstead Nov 11 '17 at 19:04
  • $\begingroup$ What is wrong with just sliding along the line y equals plus or minus x though in the original example? $\endgroup$ – J.Jones5552 Nov 11 '17 at 19:07
  • $\begingroup$ I ask because I'm really not sure how one would go about explicitly writing down a radial retraction $\endgroup$ – J.Jones5552 Nov 11 '17 at 19:10
  • $\begingroup$ @J.Jones5552: There is no continuous extension of that function to all of $I - \{ (0,0) \}$. Consider the projection onto the first coordinate; for a fixed value of $t$, the point $(x,y)$ is sent to $x$ if $|x|<|y|$ but is sent to $x+t(1-x)$ if $|x|>|y|$. This is not continuous in $x$. For example, when $t=1$ you'd be sending $(x,y)$ to $x$ when $|x|<|y|$ and to $1$ when $|x|>|y|$. $\endgroup$ – Clive Newstead Nov 11 '17 at 19:17

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