Homotopy extension property for CW pairs (Hatcher)

[ForMyThunder]

I do not understand the proof of Proposition 0.16 in Allen Hatcher's book Algebraic Topology. If someone has the book, could you please clarify the part of the proof when he says "If we perform the deformation retraction of $X^n\times I$ onto $X^n\times\{0\}\cup (X^{n-1}\cup A^n)\times I$ during the t-interval $[1/2^{n+1},\, 1/2^n]$, this infinite concatenation of homotopies is a deformation retraction of $X\times I$ onto $X\times\{0\}\cup A\times I$." I do not understand how this follows. Thanks in advance.

 

[lavinia]

I do not understand the proof of Proposition 0.16 in Allen Hatcher's book Algebraic Topology. If someone has the book, could you please clarify the part of the proof when he says "If we perform the deformation retraction of $X^n\times I$ onto $X^n\times\{0\}\cup (X^{n-1}\cup A^n)\times I$ during the t-interval $[1/2^{n+1},\, 1/2^n]$, this infinite concatenation of homotopies is a deformation retraction of $X\times I$ onto $X\times\{0\}\cup A\times I$." I do not understand how this follows. Thanks in advance.

I don't have the book but if you tll me what A^n is I will give it a shot.

 

[quasar987]

lavinia, the book is available for free on Hatcher's web page

 

[lavinia]

I do not understand the proof of Proposition 0.16 in Allen Hatcher's book Algebraic Topology. If someone has the book, could you please clarify the part of the proof when he says "If we perform the deformation retraction of $X^n\times I$ onto $X^n\times\{0\}\cup (X^{n-1}\cup A^n)\times I$ during the t-interval $[1/2^{n+1},\, 1/2^n]$, this infinite concatenation of homotopies is a deformation retraction of $X\times I$ onto $X\times\{0\}\cup A\times I$." I do not understand how this follows. Thanks in advance.

I think the Idea is that the deformation of D$^{n}$ x I

onto D$^{n}$x0 U D$^{n-1}$ X I can be followed by the cell attaching map. Over all of the n-cells this deforms X$^{n}$ onto X$^{n}$ X 0 U X$^{n-1}$ X I.

One then does the same thing on the remaining n-1 cells in X$^{n-1}$ X I and so on until you are only left with X x 0. If the complex if finite dimensional this process will stop after finitely many steps but will also work for infinite dimensional complexes such as RP$^{\infty}$

 

[blue_raver22]

The homotopy extension property for CW pairs is a fundamental property in algebraic topology that allows us to extend a homotopy defined on a subcomplex of a CW complex to the entire complex. This is a powerful tool that is used in many proofs and constructions in the field.

In Proposition 0.16 of Hatcher's book, he is proving that a CW pair (X,A) satisfies the homotopy extension property. The proof involves constructing a deformation retraction of X\times I onto X\times\{0\}\cup A\times I, which intuitively means that we can continuously deform X\times I onto X\times\{0\}\cup A\times I while keeping the points in A fixed.

The part of the proof that you are asking about involves showing that this deformation retraction can be achieved by a series of smaller deformations, each of which is defined on a smaller time interval. This is where the t-interval [1/2^{n+1},\, 1/2^n] comes into play.

Essentially, what Hatcher is saying is that if we start with a deformation retraction of X^n\times I onto X^n\times\{0\}\cup (X^{n-1}\cup A^n)\times I, and then continuously deform this onto X^{n-1}\times I, and then onto X^{n-2}\times I, and so on until we reach X\times I, we will end up with a deformation retraction of X\times I onto X\times\{0\}\cup A\times I. This is because each step in this process is a deformation retraction, and the composition of deformation retractions is also a deformation retraction.

I hope this helps clarify how the proof follows. If you are still having trouble understanding, I would suggest going through the proof step by step and trying to visualize each step in terms of the deformation retraction. Also, feel free to reach out to me for further clarification.