4

Let $A$ be the unit circle in the $xy$ plane in $3$-dimensional real space and let $A_+$ be a semicircle. I have to compute the homology of $\mathbb{R}^3\setminus A_+$. I was thinking that $\mathbb{R}^3\setminus A_+$ is homotopically equivalent to $\mathbb{R}^3\setminus \{(1,0,0)\}$ which can than be proved to be homeomorphic to $S^2$. However, I have trouble finding a retraction $g:\mathbb{R}^3\setminus \{(1,0,0)\}\rightarrow \mathbb{R}^3\setminus A_+$ to construct the equivalence, since homotopy equivalence is not stable under subtractions, i.e. if $A,B\subset X$, $A$ homotopically equivalent to $B$ it does not follow that $X\setminus A$ homotopically equivalent to $X\setminus B$.

Iulia
  • 1,306
  • It's not really clear what you mean by $A_+$. Does it include the interior of the semicircle, or just the boundary? If the latter, does it include the diameter of the circle that "chops it in half", or not? – tcamps Feb 16 '15 at 03:14
  • just the boundary. It is just something like $x^2+y^2=1, x\geq 0$. – Iulia Feb 16 '15 at 03:23
  • 1
    There is no such retraction $g$. One way to see this is to look at the composite $I \hookrightarrow \mathbb R^3 \backslash {(1,0,0)} \overset{g}{\to} \mathbb R ^3 \backslash A_+ \hookrightarrow \mathbb R^3$ where $I$ is some curve that intersects the semicircle $A_+$ at a single point $p$. Every point on $I$ other than $p$ must be fixed by this map; by continuity, $p$ must also be fixed. But since $p$ doesn't lie in $\mathbb R^3 \backslash A_+$, this is a contradiction. What is true is that the two spaces are homotopy equivalent. – tcamps Feb 17 '15 at 12:06

0 Answers0