# Exercise 0.29 Hatcher algebraic topology

Exercise 0.29 Hatcher algebraic topology. In case the CW complex $$X$$ is obtained from a subcomplex $$A$$ by attaching a single cell $$e^{n},$$ describe exactly what the extension of a homotopy $$f_{t}: A \rightarrow Y$$ to $$X$$ given by the proof of Proposition 0.16 looks like. That is, for a point $$x \in e^{n}$$, describe the path $$f_{t}(x)$$ for the extended $$f_{t}$$

Proposition 0.16. If $$(X, A)$$ is a CW pair, then $$X \times\{0\} \cup A \times I$$ is a deformation retract of $$X \times I$$, hence $$(X, A)$$ has the homotopy extension property.

Proof: There is a retraction $$r: D^{n} \times I \rightarrow D^{n} \times\{0\} \cup \partial D^{n} \times I,$$ for ex ample the radial projection from the point $$(0,2) \in D^{n} \times \mathbb{R}$$. Then setting $$r_{t}=t r+(1-t)$$ Il gives a deformation retraction of $$D^{n} \times I$$ onto $$D^{n} \times\{0\} \cup \partial D^{n} \times I .$$ This deformation retraction gives rise to a deformation retraction of $$X^{n} \times I$$ onto $$X^{n} \times\{0\} \cup\left(X^{n-1} \cup A^{n}\right) \times I$$ since $$X^{n} \times I$$ is obtained from $$X^{n} \times\{0\} \cup\left(X^{n-1} \cup A^{n}\right) \times I$$ by attach ing copies of $$D^{n} \times I$$ along $$D^{n} \times\{0\} \cup \partial D^{n} \times I .$$ If we perform the deformation retrac tion of $$X^{n} \times I$$ onto $$X^{n} \times\{0\} \cup\left(X^{n-1} \cup A^{n}\right) \times I$$ during the $$t$$ -interval $$\left[1 / 2^{n+1}, 1 / 2^{n}\right]$$ this infinite concatenation of homotopies is a deformation retraction of $$X \times I$$ onto $$X \times\{0\} \cup A \times I .$$ There is no problem with continuity of this deformation retraction at $$t=0$$ since it is continuous on $$X^{n} \times I$$, being stationary there during the $$t$$ -interval $$\left[0,1 / 2^{n+1}\right],$$ and $$\mathrm{CW}$$ complexes have the weak topology with respect to their skeleta so a map is continuous iff its restriction to each skeleton is continuous.

• Is there a question in here that I missed somehow? May 16, 2020 at 11:23
• @Tyrone . describe exactly what the extension of a homotopy $f_{t}: A \rightarrow Y$ to $X$....?
– user768579
May 16, 2020 at 11:35
• The question mark is a start, but I'm still confused as to what you expect a good answer to include. Try being more specific. May 16, 2020 at 11:49

I am also trying to solve this problem; here is what I wrote down. The key is to check out the illustration for the proof of Proposition 0.16. In order to visualize the proof, we mostly look at the case where $$n = 2$$, but the approach on higher cells can also be generalized.

We first study the proof of proposition 0.16. The proof involves a toy example, where the space $$X = D$$ is just a disk, and the space $$A = \partial D$$, then proposition 0.16 says that there is a deformation retraction from $$D\times I$$ to $$D\times \{0\}\cup \partial D\times I$$. Therefore, we are looking for a way to deform retract the cylinder to the glass-cup shape colored in yellow. The way it worked in Hatcher is that we find a point above the cylinder, and for any point of the cylinder, we perform a projection so that it lands either on the wall or on the bottom of the cylinder.

This toy example would help with out intuition in what comes next. Say we have a map $$F:X\to Y$$, and let $$X = A\cup \{e^n\}$$. We know there is a family of maps $$f_t:A\times I \to Y$$ as a homotopy, given as the restriction of $$F$$ onto $$A$$. We now need to construct a homotopy $$\bar f_t: X\times I\to Y$$ such that

• it restricts to $$f_t:A\times I\to Y$$ when we are on $$A\subseteq X$$, and
• it restricts to $$F:X\to Y$$ at time $$t = 0$$.

We first visualize the scenario. We denote $$A$$ above to be the (boundary of) subcomplex we are working on, and let the disk $$e$$ be the attachment we have onto $$A$$.

We want to show that there is a homotopy from this connected space to some unknown space $$Y$$. We have already know how the map works on $$A\times [0,1]$$, which is the region covered in red; we also know how the map works on $$e\times \{0\}$$, which is the space covered in yellow. The issue at hand is to find a way and describe the mapping on the attached cylinder below in the figure below.

Therefore, we just want to know how the mapping is done for any point of the cylinder not on $$A\times I$$ nor on $$e\times \{0\}$$. However, we realize this is essentially a cylinder and therefore our toy example can help us. Via that example, we know by identifying some point above the cylinder, we can project every point of the cylinder to the wall and the bottom. Therefore, the problem has been reduced to defining the pathway on the outer wall. But this is just a $$1$$-dimensional version of the toy example we have: if we put down the wall flat like a piece of paper, then by identifying some point above it, there is a projection so that every point either lands on the side boundary or the bottom boundary. At this point, we are done: the side boundary corresponds to points on $$A\times [0,1]$$, and the bottom boundary corresponds to points on $$e\times \{0\}$$, and either way this is already defined.

In conclusion, given a point $$x\in e^n$$ as mentioned, the path $$f_t(x)$$ consists of possibly a few steps:

• Identify some point $$*$$ directly above the outer arc $$e^n$$, and we just have to define the mapping of $$e^n\times [0,1]$$ to the space $$Y$$.
• Given any point in $$e^n\times [0,1]$$, we project them to the boundary $$e^n\times \{0\}\cup \partial e^n\times I$$.
• Similar above, we do another projection so that for any point in $$e^n\times \{0\}\cup \partial e^n\times I$$, it ends up on $$e^n\times \{0\}\cup \partial (\partial e^n)\times I$$.
• Continue inductively as necessary, but eventually all points should land in somewhere defined. The mapping of our original point will depend on this point we obtain after projection.

Let $$f:A\times I\rightarrow Y$$ be the homotopy and $$F_0:X\rightarrow Y$$ an extension of $$f_0$$. The extension you're looking for is gotten by retracting $$X\times I$$ to $$X\times\{0\}\cup A\times I$$ and then applying either $$F_0$$ or $$f$$. This is all pretty explicit with the retraction given in the proof of 0.16; each point of the disk is sliding radially towards the boundary, and then once it gets there, sliding up $$A\times I$$.

If $$\Phi:\mathbf{D}^n\rightarrow X$$ is the characteristic map for $$e_n$$, then we can write this map out, with $$z\in\mathbf{D}^n$$: $$\tilde{F}(z,t) = \begin{cases} F_0\Phi\big(\frac{2z}{2-t}\big) & |z|\le\frac{2-t}{2}\\ f\big(\Phi(\frac{z}{|z|}), 2-\frac{2-t}{|z|}\big)&\text{else} \end{cases}$$

Then it's easy to see $$\tilde{F}$$ and $$f$$ piece together to give the necessary extension $$F:X\times I\rightarrow Y$$.