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

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?)

Thank you.

• 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 – Justin Young May 17 '19 at 15:19
• Maybe 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. – Justin Young May 17 '19 at 15:44
• Ok, 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. – Justin Young May 22 '19 at 17:46
• 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. – Justin Young May 23 '19 at 20:14
• @JustinYoung Thanks a lot for your comments and for tracking down that reference. If you post it as an answer I will accept it... but bounty ends in 1 hour! – Nasos Evangelou-Oost May 24 '19 at 4:45

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 $$M\to K$$.

• The answer below (of Mathy) shows that the tubular neighborhood is a (countable) cell complex, and by general homotopy theory, this is homotopy equivalent to a CW complex. This answers $3$. – Justin Young May 31 '19 at 15:07

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.

• This works well if $M$ is compact, but if $M$ is not compact the tube might have arbitrarily small diameter, and then $1/n$ is too big to "fill the open set up". Could you elaborate how this works with details? – Justin Young May 28 '19 at 11:25
• I am not sure what you mean, but by the following argument you can see that any open subset $U$ of $\mathbb{R}^n$ is a CW-complex: The idea is to create a grid of $\mathbb{R}^n$ of grid-edge-lengths all equal to $1/n$ and vertices therefore all rational. Then you take those boxes of the grid, which lie completely in $U$. Do that for all $n$ and you covered $U$ completely with boxes of rational edge lengths and rational vertices, so that there are only countably many (do we need countability for CW-complexes?). This gives $U$ the structure of a CW-complex (each box is a $D^n$). – Mathy May 28 '19 at 16:07
• Ok, I'm convinced you have some kind of cell complex there, to get a true CW structure requires some work, unless you can tell me precisely what are the cells and how do the attachments work? – Justin Young May 28 '19 at 18:15
• @Mathy thanks for your comments. Do you know of a written reference for the claim that every open subset of $R^n$ is a CW complex? It does sound plausible but I'd like to see some more details as well. Thanks again. – Nasos Evangelou-Oost May 29 '19 at 11:37
• Unfortunately I don't have any reference either. @Justin: I can tell you that precisely. Once you've covered your whole open subset $U \subset \mathbb{R}^n$ up with those boxes (and we can even forget about the rationality arguments, because a CW-complex can also have uncountably many cells), you take the grid points as the 0-skeleton $U_0$. Then for every box, your attaching map $S^n \to U_0$ is given through connecting the corresponding grid points of the box via straight lines (one can use the homeomorphism $S^n \to \partial I^n$). – Mathy May 29 '19 at 17:10