I am reading the section about CW approximations in Hatcher's algebraic topology (the 2015 reprint), and I am having trouble with understanding the proof of uniqueness for CW approximation (up to homotopy equivalence). His argument goes as follows:
If $f:Z\rightarrow X, g:Z'\rightarrow X$ are both CW approximations for $X$, we can look at the composition $Z\stackrel{f}{\rightarrow} X \hookrightarrow M_g$, where $M_g$ is the mapping cylinder of $g$ and from what I understand the second map is the inclusion of $X$ into the mapping cylinder.
He then claims that by the compression lemma, the composition above can be deformed into $Z'$ which gives a weak homotopy equivalence between $Z$ and $Z'$ (and hence a homotopy equivalence, due to Whitehead's theorem).
I am not sure about two things:
How is the compression lemma applied? I understand that we consider $Z'$ as a subspace of the mapping cylinder, but I do not understand what subspace of $Z$ we are considering in the domain and why the conditions for the lemma are met.
Why does the composition give a weak homotopy equivalence? The induced map of homotopy groups is certainly injective as the composition of injections. But why is is surjective? The first arrow is surjective, and the second arrow is probably surjective somehow due to how we use the compression lemma, but I do not understand why (which relates to my first question).
Can somebody please explain this to me?
Thanks in advance.
I have thought about this, and this is the explanation I came up with:
The compression lemma is applied to the pairs $(Z,z_0)\rightarrow (M_f,Z')$.
Since $Z'$ is a deformation retract of $M_f$, the conditions for the compression lemma are met automatically.
The second induced map in the homotopy groups (the right arrow) is surjective since in this particular case, it is easy to see (via, say, the long exact sequence of homotopy groups) that $\pi_n(M_f,X)=0$ for all $n$, and so the inclusion induces an isomorphism.