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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTime7 days ago
    • (edited 7 days ago)

    At CW-complexes are paracompact Hausdorff spaces

    I wrote out proof of the lemma that the result of attaching a cell to a paracompact Hausdorff space is still paracompact Hausdorff (here).

    Not very nice yet. Needs polishing and maybe some more general lemmas.

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTime5 days ago
    • (edited 5 days ago)

    Did some further polishing. It’s still not a fun read, but there is now

    1. lemma and proof that single cell attachment to a Hausdorff space is still Hausdorff (here)

    2. lemma and proof that single cell attachment to a paracompact Hausdorff space is still paracompact Hausdorff (here)

    So there is now proof spelled out that finite CW-complexes (and finite relative CW-complexes relative to paracompact Hausdorff spaces) are paracompact Hausdorff.

    I think I’ll refrain from doing the general case, unless somebody points out a really elegant way to do it.

    • CommentRowNumber3.
    • CommentAuthorTodd_Trimble
    • CommentTime5 days ago

    In the proof of the first claim of Lemma 1.3, item 1.: to see that i D n:D nX fD ni_{D^n}: D^n \to X \cup_f D^n is a closed map, can’t you just use the fact that maps from a compact space like D nD^n to a Hausdorff space like X fD nX \cup_f D^n are automatically closed?

    If it is of any interest, item 2. was also treated here.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTime5 days ago
    • (edited 5 days ago)

    Thanks, Todd!

    use the fact that maps from a compact space like D nD^n to a Hausdorff space like X fD nX \cup_f D^n are automatically closed?

    True, I should appeal to maps from compact spaces to Hausdorff spaces are closed and proper.

    In fact for a while I thought I could give a nice proof of Hausdorffness by first forming the one-point compactification of the paracompact Hausdorff XX, then attach to that compactification, pick the evident subspace in the result and deduce Hausdorffness by appealing to quotient projections out of compact Hausdorff spaces are closed precisely if the codomain is Hausdorff.

    Why did I abandon this? Now I forget. Maybe I shouldn’t have.

    If it is of any interest, item 2. was also treated here.

    Thanks, for the pointer. That’s nice, I’ll add a reference to that.

    • CommentRowNumber5.
    • CommentAuthorTodd_Trimble
    • CommentTime4 days ago

    Here is a germ of an idea for an alternative approach.

    Lemma: If XX is T 4T_4 and EE is a closed equivalence relation, then X/EX/E is also T 4T_4 (and in particular, Hausdorff). Proof: I read this statement somewhere on the internet. Seriously: I don’t imagine this is hard, but I haven’t written out a proof myself. It would be a handy statement to have at the nLab.

    Now, if we assume (as part of an inductive argument) XX is paracompact Hausdorff, then by Dieudonne’s theorem it is T 4T_4, and then the attachment space X fD nX \cup_f D^n is also T 4T_4 by the preceding lemma. I guess we’d have to prove that the equivalence relation on XD nX \sqcup D^n, generated by xyx \sim y if f(x)=yf(x) = y for xS n1x \in S^{n-1} and yXy \in X, actually is closed. Again, I don’t imagine that’s hard and I imagine there’s an abstract proof for it, but I haven’t written it out.

    But that would give the Hausdorffness of X fD nX \cup_f D^n whose current proof is a bit fiddly. The paracompactness of X fD nX \cup_f D^n seems well in hand.

    In the transfinite induction proof, this at least gives the inductive step at successor stages. Then it would be nice to have a statement like this:

    Theorem: If ITop:iX iI \to Top: i \mapsto X_i is an limit ordinal chain of paracompact Hausdorff spaces all of whose transition maps are closed, then the colimit XX is also paracompact Hausdorff (and each map X iXX_i \to X is also closed).

    The bit about XX being Hausdorff is probably easy (and maybe even a corollary of the Lemma above). Not sure about the bit about XX being paracompact. The last bit about X iXX_i \to X being closed is easy.

    • CommentRowNumber6.
    • CommentAuthorTodd_Trimble
    • CommentTime4 days ago

    (Actually, I misread that thing on the internet, so those suggestions probably need to cook a little longer. Sorry for noise.)

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTime4 days ago
    • (edited 4 days ago)

    The paracompactness of X fD nX \cup_f D^n seems well in hand.

    Yes, I liked that part of the argument. But I just noticed a gap: For invoking that lemma on intermediate saturated subsets, I need the closed subset to already be saturated itself. I had overlooked this. For i Xi_X this is automatic, but for i D ni_{D^n} this will fail to the extent that ff is not injective.

    Grr. And I need to quit now.

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTime4 days ago

    This is maybe trickier than it might seem. The short proof in Hatcher is just a sketch, or else I am missing something. The proof in Lee is tedious but works after some fixes (here).

    • CommentRowNumber9.
    • CommentAuthorTodd_Trimble
    • CommentTime4 days ago

    I’ll try to take a closer look at paracompactness sometime soon.

    But I would like to take up the business about Hausdorffness again, and try a different arrangement which proves a more general result, with intermediate results which I think are of independent interest. (I must confess, I didn’t quite follow what you wrote for item 1 in the proof of Lemma 1.3, the part coming after “More explicitly”. So I wanted to start over from scratch.)

    Proposition 1: Let XX be a T 4T_4 space, and let YY be a quotient space of XX such that the quotient map q:XYq: X \to Y is closed. Then YY is also T 4T_4.

    Proof: First we claim singletons of YY are closed. Proof: for each yYy \in Y there exists xXx \in X such that y=q(x)y = q(x). Since {x}\{x\} is closed in XX and qq is closed, {y}=q({x})\{y\} = q(\{x\}) is closed in YY.

    Let C,DC, D be disjoint closed sets of YY, so that ¬C,¬D\neg C, \neg D form an open cover of YY. Then A=q 1(¬C),B=q 1(¬D)A = q^{-1}(\neg C), B = q^{-1}(\neg D) form an open cover of XX. Normality of XX (in “De Morganized” form) implies there are closed subsets EA,FBE \subseteq A, F \subseteq B which together also cover XX. Then Y=q(X)=q(EF)=q(E)q(F)Y = q(X) = q(E \cup F) = q(E) \cup q(F), so the closed sets q(E),q(F)q(E), q(F) cover YY. And we have q(E)q(A)=¬Cq(E) \subseteq q(A) = \neg C and similarly q(F)¬Dq(F) \subseteq \neg D, which is equivalent to C¬q(E)C \subseteq \neg q(E) and D¬q(F)D \subseteq \neg q(F), where ¬q(E),¬q(F)\neg q(E), \neg q(F) are disjoint open sets of YY, and we are done. \Box

    Lemma 2: If XX is a compact Hausdorff space and X×X\sim \hookrightarrow X \times X is an equivalence relation which is closed as a subset of X×XX \times X, then the quotient space X/X/\sim is also compact Hausdorff. (Consequently, the quotient map q:XX/q: X \to X/\sim is closed.)

    Proof: omitted for now. \Box

    Proposition 3: Given a pushout diagram in TopTop

    X h Z f g Y k W,\array{ X & \stackrel{h}{\to} & Z \\ \mathllap{f} \downarrow & & \downarrow \mathrlap{g} \\ Y & \underset{k}{\to} & W, }

    if hh is monic, ff is epic, and X,Y,ZX, Y, Z are compact Hausdorff, then WW is compact Hausdorff.

    Proof: The kernel pair of ff is a closed equivalence relation EE on XX such that ff is the coequalizer of the first and second pullback projections p 1,p 2:EXp_1, p_2: E \rightrightarrows X. The quotient map g:ZWg: Z \to W is similarly the quotient of the closed equivalence relation on ZZ formed as (h×h)(E)Δ Z(h \times h)(E) \cup \Delta_Z. Now apply the preceding lemma. \Box

    Lemma 4: The pushout of a closed embedding along any continuous map is again a closed embedding.

    Proof: omitted for now, but we’ve covered this before. \Box

    Proposition 5: Let h:XZh: X \to Z be an embedding of compact Hausdorff spaces, and let f:XYf: X \to Y be a continuous map into a Hausdorff space. Then in the pushout in TopTop

    X h Z f g Y k W\array{ X & \stackrel{h}{\to} & Z \\ \mathllap{f} \downarrow & & \downarrow \mathrlap{g} \\ Y & \underset{k}{\to} & W }

    the map gg is closed.

    Proof: Factorize f:XYf: X \to Y into an epi e:XKe: X \to K followed by a mono m:KYm: K \to Y. Then we have a composition of pushout squares in TopTop

    X h Z e e K j K m m Y k W\array{ X & \stackrel{h}{\to} & Z \\ \mathllap{e} \downarrow & & \downarrow \mathrlap{e'} \\ K & \stackrel{j}{\to} & K' \\ \mathllap{m} \downarrow & & \downarrow \mathrlap{m'} \\ Y & \underset{k}{\to} & W }

    where KK' is compact Hausdorff by Proposition 3, so that ee' is closed. Also mm is a closed embedding, so mm' is closed by Lemma 4. Hence g=meg = m'e' is closed. \Box

    Proposition 6: Let XX be paracompact and Hausdorff, and let f:S n1Xf: S^{n-1} \to X be a continuous map. Then the attachment space X fD nX \cup_f D^n is Hausdorff.

    Proof: The space XX is T 4T_4 by Dieudonne’s theorem, so XD nX \cup D^n is also T 4T_4, so by Proposition 1 it suffices to check that the quotient map XD nX fD nX \cup D^n \to X \cup_f D^n is closed, or equivalently that each of the pushout coprojections XX fD nX \to X \cup_f D^n and D nX fD nD^n \to X \cup_f D^n is closed. The embedding XX fD nX \to X \cup_f D^n, being a pushout of the closed embedding S n1D nS^{n-1} \hookrightarrow D^n along the continuous map ff, is closed by Lemma 4. The map D nX fD nD^n \to X \cup_f D^n is closed by Proposition 5. \Box

    • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTime4 days ago

    Hi Todd,

    thanks, that looks good. In particular your argument seems to extend to the case that we are attaching not just a single cell but a whole set of cells at once, which is what we should really be after.

    I’ll stay away from this entry now for a bit, need to do look into some other things. So please feel invited to add this material there.

    This morning I (believe I) fixed the gap mentioned in #7. Now that I am awake again this is obvious enough: apply the previous argument just to the complements D nS n1D^n \setminus S^{n-1} and Xf(S n1)X \setminus f(S^{n-1}) to obtain locally finite refinements of the original cover on these subspaces, then invoke the compactness of f(S n1)f(S^{n-1}) to throw in a remaining finite set of patches covering f(S n1)f(S^{n-1}) in X fS n1X \cup_f S^{n-1}. The end result is a locally finite cover of X fD nX \cup_f D^n.

    Unfortunately this argument does not seem to generalize to attachment of a non-finite set of cells.