9
$\begingroup$

This is regarding the fundamental group computation of the complement of a toral knot in $S^3$ in Hatcher's algebraic topology book. See page 48. I have understood till the stage where the cross section of the torus minus the knot deformation-retracts to the radial segments as the arrows indicate. What is not clear is "Letting $x$ vary, these radial segments then trace out a copy of the mapping cylinder $X_m$ in the first solid torus."

I tried imagining this with a simple cases like the trefoil knot, but can't fathom this statement. Any help would be greatly appreciated!

$\endgroup$
2
  • 7
    $\begingroup$ It helps to think of the 3 sphere as the union of two solid tori. The circles hatcher uses in the mapping cylinder construction are the cores of the solid torus plus a curve parallel to the knot in the separatingtorus. $\endgroup$ Jul 11, 2012 at 16:44
  • 1
    $\begingroup$ Never thought of it this way! Will chew upon this. Thanks a ton! $\endgroup$
    – Karthik C
    Jul 11, 2012 at 16:58

2 Answers 2

9
$\begingroup$

It helps to think of the 3 sphere as the union of two solid tori. The circles Hatcher uses in the mapping cylinder construction are the cores of the solid torus plus a curve parallel to the knot in the separating torus.

$\endgroup$
2
  • $\begingroup$ If this is the case, then even if each torus deformation retracts onto each of these cylinders it doesn’t deformation retract onto X_mn, the original required space, since the two « cylinders » don’t intersect in the required circle no? $\endgroup$ Feb 13 at 19:54
  • $\begingroup$ @LittleNarwhal the two cylinders should intersect on the curve parallel to the knot in the separating torus - which is a copy of $S^1$. This is also the 'domain' of either mapping cylinder, so it is precisely what deformation retracts onto the respective ends. $\endgroup$ Mar 11 at 14:42
0
$\begingroup$

To expound a bit:

The center of the "Y" figure is a point of $S^1$ of the mapping cylinder. This center also wraps around the torus once as $x$ varies. However, the tips of the "Y" wrap around the torus $m$ times.

$\endgroup$

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.