0
$\begingroup$

Hatcher's Algebraic Topology says the following on pg. 2

Not all deformation retractions arise in this way from mapping cylinders, however. For example, the thick $\mathbf{X}$ deformation retracts to the thin $X$, which in terms retracts to the point of intersection of its two crossbars. The net result is a deformation retraction of $\mathbf{X}$ onto a point, during which certain pairs of points follow paths that merge before reaching their final destination.

I don't understand this. Why can't the two cross bars retract to the point of intersection without any two paths crossing?

$\endgroup$
1
$\begingroup$

Hatcher is stating a property that this particular retraction mapping has; he is not stating a property that all retraction mappings would have. He is simply describing how it 'works' in this instance, what it looks like.

$\endgroup$
2
  • $\begingroup$ I am talking about this instance too. The way I can visualize it, each point on $X$ would deform to a single point without any two paths crossing. $\endgroup$ – algebraically_speaking Feb 5 '15 at 7:05
  • $\begingroup$ You can certainly do it that way. There is not a problem with that. In Hatcher's example he's simply doing it another way, where the thick $\mathbf{X}$ first retracts onto the thin $X$. There's no conflict, because it's just one example of a retraction. $\endgroup$ – Michael Cromer Feb 5 '15 at 7:14

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.