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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.
Did some further polishing. It’s still not a fun read, but there is now
lemma and proof that single cell attachment to a Hausdorff space is still Hausdorff (here)
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.
In the proof of the first claim of Lemma 1.3, item 1.: to see that $i_{D^n}: D^n \to X \cup_f D^n$ is a closed map, can’t you just use the fact that maps from a compact space like $D^n$ to a Hausdorff space like $X \cup_f D^n$ are automatically closed?
If it is of any interest, item 2. was also treated here.
Thanks, Todd!
use the fact that maps from a compact space like $D^n$ to a Hausdorff space like $X \cup_f D^n$ 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 $X$, 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.
Here is a germ of an idea for an alternative approach.
Lemma: If $X$ is $T_4$ and $E$ is a closed equivalence relation, then $X/E$ is also $T_4$ (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) $X$ is paracompact Hausdorff, then by Dieudonne’s theorem it is $T_4$, and then the attachment space $X \cup_f D^n$ is also $T_4$ by the preceding lemma. I guess we’d have to prove that the equivalence relation on $X \sqcup D^n$, generated by $x \sim y$ if $f(x) = y$ for $x \in S^{n-1}$ and $y \in X$, 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 $X \cup_f D^n$ whose current proof is a bit fiddly. The paracompactness of $X \cup_f D^n$ 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 $I \to Top: i \mapsto X_i$ is an limit ordinal chain of paracompact Hausdorff spaces all of whose transition maps are closed, then the colimit $X$ is also paracompact Hausdorff (and each map $X_i \to X$ is also closed).
The bit about $X$ being Hausdorff is probably easy (and maybe even a corollary of the Lemma above). Not sure about the bit about $X$ being paracompact. The last bit about $X_i \to X$ being closed is easy.
(Actually, I misread that thing on the internet, so those suggestions probably need to cook a little longer. Sorry for noise.)
The paracompactness of $X \cup_f D^n$ 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 $i_X$ this is automatic, but for $i_{D^n}$ this will fail to the extent that $f$ is not injective.
Grr. And I need to quit now.
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).
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 $X$ be a $T_4$ space, and let $Y$ be a quotient space of $X$ such that the quotient map $q: X \to Y$ is closed. Then $Y$ is also $T_4$.
Proof: First we claim singletons of $Y$ are closed. Proof: for each $y \in Y$ there exists $x \in X$ such that $y = q(x)$. Since $\{x\}$ is closed in $X$ and $q$ is closed, $\{y\} = q(\{x\})$ is closed in $Y$.
Let $C, D$ be disjoint closed sets of $Y$, so that $\neg C, \neg D$ form an open cover of $Y$. Then $A = q^{-1}(\neg C), B = q^{-1}(\neg D)$ form an open cover of $X$. Normality of $X$ (in “De Morganized” form) implies there are closed subsets $E \subseteq A, F \subseteq B$ which together also cover $X$. Then $Y = q(X) = q(E \cup F) = q(E) \cup q(F)$, so the closed sets $q(E), q(F)$ cover $Y$. And we have $q(E) \subseteq q(A) = \neg C$ and similarly $q(F) \subseteq \neg D$, which is equivalent to $C \subseteq \neg q(E)$ and $D \subseteq \neg q(F)$, where $\neg q(E), \neg q(F)$ are disjoint open sets of $Y$, and we are done. $\Box$
Lemma 2: If $X$ is a compact Hausdorff space and $\sim \hookrightarrow X \times X$ is an equivalence relation which is closed as a subset of $X \times X$, then the quotient space $X/\sim$ is also compact Hausdorff. (Consequently, the quotient map $q: X \to X/\sim$ is closed.)
Proof: omitted for now. $\Box$
Proposition 3: Given a pushout diagram in $Top$
$\array{ X & \stackrel{h}{\to} & Z \\ \mathllap{f} \downarrow & & \downarrow \mathrlap{g} \\ Y & \underset{k}{\to} & W, }$if $h$ is monic, $f$ is epic, and $X, Y, Z$ are compact Hausdorff, then $W$ is compact Hausdorff.
Proof: The kernel pair of $f$ is a closed equivalence relation $E$ on $X$ such that $f$ is the coequalizer of the first and second pullback projections $p_1, p_2: E \rightrightarrows X$. The quotient map $g: Z \to W$ is similarly the quotient of the closed equivalence relation on $Z$ formed as $(h \times h)(E) \cup \Delta_Z$. 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 $h: X \to Z$ be an embedding of compact Hausdorff spaces, and let $f: X \to Y$ be a continuous map into a Hausdorff space. Then in the pushout in $Top$
$\array{ X & \stackrel{h}{\to} & Z \\ \mathllap{f} \downarrow & & \downarrow \mathrlap{g} \\ Y & \underset{k}{\to} & W }$the map $g$ is closed.
Proof: Factorize $f: X \to Y$ into an epi $e: X \to K$ followed by a mono $m: K \to Y$. Then we have a composition of pushout squares in $Top$
$\array{ X & \stackrel{h}{\to} & Z \\ \mathllap{e} \downarrow & & \downarrow \mathrlap{e'} \\ K & \stackrel{j}{\to} & K' \\ \mathllap{m} \downarrow & & \downarrow \mathrlap{m'} \\ Y & \underset{k}{\to} & W }$where $K'$ is compact Hausdorff by Proposition 3, so that $e'$ is closed. Also $m$ is a closed embedding, so $m'$ is closed by Lemma 4. Hence $g = m'e'$ is closed. $\Box$
Proposition 6: Let $X$ be paracompact and Hausdorff, and let $f: S^{n-1} \to X$ be a continuous map. Then the attachment space $X \cup_f D^n$ is Hausdorff.
Proof: The space $X$ is $T_4$ by Dieudonne’s theorem, so $X \cup D^n$ is also $T_4$, so by Proposition 1 it suffices to check that the quotient map $X \cup D^n \to X \cup_f D^n$ is closed, or equivalently that each of the pushout coprojections $X \to X \cup_f D^n$ and $D^n \to X \cup_f D^n$ is closed. The embedding $X \to X \cup_f D^n$, being a pushout of the closed embedding $S^{n-1} \hookrightarrow D^n$ along the continuous map $f$, is closed by Lemma 4. The map $D^n \to X \cup_f D^n$ is closed by Proposition 5. $\Box$
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 $D^n \setminus S^{n-1}$ and $X \setminus f(S^{n-1})$ to obtain locally finite refinements of the original cover on these subspaces, then invoke the compactness of $f(S^{n-1})$ to throw in a remaining finite set of patches covering $f(S^{n-1})$ in $X \cup_f S^{n-1}$. The end result is a locally finite cover of $X \cup_f D^n$.
Unfortunately this argument does not seem to generalize to attachment of a non-finite set of cells.
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