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

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    $\begingroup$ 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 $\endgroup$ – Justin Young May 17 '19 at 15:19
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    $\begingroup$ 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. $\endgroup$ – Justin Young May 17 '19 at 15:44
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    $\begingroup$ 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. $\endgroup$ – Justin Young May 22 '19 at 17:46
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    $\begingroup$ 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. $\endgroup$ – Justin Young May 23 '19 at 20:14
  • $\begingroup$ @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! $\endgroup$ – Nasos Evangelou-Oost May 24 '19 at 4:45
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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 $M\to K$.

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  • $\begingroup$ 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 $3$. $\endgroup$ – Justin Young May 31 '19 at 15:07
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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.

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    $\begingroup$ 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? $\endgroup$ – Justin Young May 28 '19 at 11:25
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    $\begingroup$ 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 $D^n$). $\endgroup$ – Mathy May 28 '19 at 16:07
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    $\begingroup$ 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? $\endgroup$ – Justin Young May 28 '19 at 18:15
  • $\begingroup$ @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. $\endgroup$ – Nasos Evangelou-Oost May 29 '19 at 11:37
  • $\begingroup$ 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 $S^n \to \partial I^n$). $\endgroup$ – Mathy May 29 '19 at 17:10

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