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$^{\infty}$