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?


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.

  • $\begingroup$ 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

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