Milnor's proof that a smooth manifold has the homotopy type of a CW complex

Question

I have some questions about the proof of Theorem 3.5 of Milnor’s “Morse Theory”: At the end of the proof of this theorem, Milnor addresses the case when $f$ has infinitely many critical points: Questions:

1. Is the limit map $g$ the colimit of the graphs of the maps $g_i : M^{a_i} \to K_i$?, i.e. $\textrm{Graph}(g_i) = \{(g_i(m), m) : m \in M^{a_i}\}$ (as each map extends the previous one, the colimit of the graphs (i.e. a colimit of sets) makes sense.)

2. Why does $g$ induce isomorphisms of homotopy groups in all dimensions?

3. Why does $M$ being a retract of its tubular neighbourhood in a euclidean space mean it is dominated by a CW complex? (Hatcher, Proposition A.11 p. 528. has an elementary proof that "A space dominated by a CW complex is homotopy equivalent to a CW complex.", so if it is obvious that $M$ is dominated by a CW complex, why was Morse theory even needed to prove Theorem 3.5?)

Google books has the section available to view (p23-24) https://books.google.com.au/books?id=A9QZZ3S_QxwC&lpg=PA25&pg=PA23#v=onepage&q&f=false

Thank you.

Answer 1

Posting comments as an answer, response to question in comments below.

On 1 and 2, yes "limit" is old-school terminology for "colimit". Under nice assumptions, the colimit of a sequence of homotopy equivalences is a weak equivalence, see for example: lemma on p. 67 of May math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf

On the parenthetical comment in 3: the important point of the Morse theory is not that the space has the homotopy type of a CW complex, but that you can get its cell structure from Morse theory.

On 3: I am convinced that Milnor is hiding some gory details. I have not found a reasonable proof that the tubular neighborhood gives a dominating CW-complex. Every standard reference I could find deals only with the case that M is compact. In this case, 3 is trivial. There are some technical proofs out there that an ENR is dominated by a CW-complex (old 40s stuff). There are also methods of triangulation of smooth manifolds. I really don't know what Milnor had in mind specifically, but, I think you'll have to dig into the literature to get a full answer. A starting point would be Milnor, On spaces having the homotopy type of a CW-complex. In there, you'll see that he dodges exactly the issue raised by 3. The reference he gives is: O. Hanner, Some theorems on absolute neighborhood retracts, Ark. Mat. vol. 1 (1950) pp.389-408.

Finally, if we look at May's Lemma, first of all it should be clear that the isomorphism $\text{colim } \pi_n(X_i) \cong \pi_n(X)$ is not just any old isomorphism, but it is induced by the maps $X_i\to X$. Now, if we follow your chain of isomorphisms above, you see that the isomorphism $\text{colim } \pi_n(M^{a_i}) \cong \text{colim } \pi_n(K_i)$ is induced by the maps $M^{a_i} \to K_i$. Since the map $M\to K$ is the colimit map (see above), it is induced by $M^{a_i}\to K_i$, and therefore the isomorphism $\pi_n(M) \cong \text{colim } \pi_n(M^{a_i}) \cong \text{colim } \pi_n(K_i) \cong \pi_n(K)$ is induced by the colimit map $M\to K$.

Answer 2

Ad 3) $M$ can be embedded in $\mathbb{R}^N$ for $N \in \mathbb{N}$ by Whitney. Then, a tubular neighbourhood of $M$ in $\mathbb{R}^N$ is an open subset of $\mathbb{R}^N$. And open subsets of $\mathbb{R}^N$ are CW-complexes (choose a grid of edge length $1/n$ to fill the open set up). Therefore $M$ is dominated by a CW-complex per definitionem. Finally, as you mentioned A.11 in Hatcher and Whitehead's theorem complete the proof.