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I have two small questions related to some point-set topology involved in the following. My question is inspired after reading the proof of Proposition 1.26 of Hatcher.

Suppose I have a space $X$ that is built out of a subspace $A$ that is path-connected by attaching some $n$ - cells $e_\alpha^n$ for $n \geq 3$ via attaching maps $\varphi_\alpha : S^{n-1} \to A$. Suppose for each $\alpha$ I choose a point $y_\alpha \in D_\alpha$. Now suppose I set

$$U = X - \left\{\bigcup_{\alpha} y_\alpha\right\}.$$

I am interested to know why $(1)$ $U$ is open in $X$. It could be that a union of infinitely many point sets is not closed, from which we cannot conclude that $U$ is open. How can I see this fact?

My second question is somewhat related to the proposition as well in that I believe:

$U$ is homotopy equivalent to my subspace $A$ from which $X$ is built out of.

Now I don't know if this is true and I am trying to prove it. I already have an inclusion map $i : A \to U$ to use. However, I don't necessarily have a map going the other way from $U$ to $A$. Is my belief true and if it is, how would I go along proving it? At the end of the day, I would like to be able to fill in the details of this by myself.

Thanks.

Edit: I should say that my $\alpha$'s run over an arbitrary index set, not necessarily countable or anything.

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What are your $D_\alpha$? Closed or open cells? If they're closed, the answer to your second question is no. –  Miha Habič Aug 13 '12 at 23:58
    
@MihaHabič My $D_\alpha$ a priori were closed disks. –  fpqc Aug 14 '12 at 0:00
    
@MihaHabič What goes wrong if they are closed disks? –  fpqc Aug 14 '12 at 0:13
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1 Answer

up vote 2 down vote accepted

I'm first going to answer the question I think you meant to ask. If I'm mistaken, please say so and I'll remove this.

Looking at the section in Hatcher you indicated, I think you meant your $D_\alpha$ to be open cells, i.e. images of open disks with the characteristic maps (this is what Hatcher denotes $e^n_\alpha$). In this case it is simple to see that $U$ is open. Recall that a subset of a CW complex is open/closed iff it intersects each closed cell in an open/closed set (this is a defining characteristic of cell complexes; Hatcher discusses this in the appendix, I think). The intersection of your set $U$ and a closed cell in $X$ is simply the cell, minus an interior point. Pulling back to the disk, this is clearly open, so $U$ must be open.

Similarly, if your $D_\alpha$ are open cells, the answer to your second question is yes. To prove this, it suffices to show that $U$ deformation retracts onto $A$. Let me give a quick sketch of this in the case there is only one (aka finitely many) cell(s). We have removed a point from the interior of the single cell. Pulling this cell back, its preimage with its characteristic map is a punctured disk (remember that the characteristic map is a homeomorphism when restricted to the interior of the disk). There is an obvious deformation retraction of this onto the boundary $S^1$. Now just take this deformation retraction and combine it with the characteristic map to get a deformation retraction of the (closed) cell onto its boundary in $X$. Since this boundary lies in $A$, we are done.

As I said, the same works for finitely many cells. When dealing with infinitely many cells, you might need to use a (fairly standard) contrivance to be able to contract all of the cells in "finite time". This method is also shown in Hatcher, probably somewhere around the point where he introduces $S^\infty$.


If you insist on having $D_\alpha$ be closed cells, I expect things can go wrong. I think $U$ might not be open in this case, but I'm too rusty at the moment to think of a counterexample.

Certainly what you conjecture in your second question is false in this case. Take a closed interval as $A$ and attach an $n$-cell at one of the boundary points. Let that boundary point be the point $y$ you remove. Then $U$ isn't even connected and can't be homotopy equivalent to $A$.

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The thing to remember is that the interiors of cells behave nicely in a CW complex. The boundaries can get weird. –  Miha Habič Aug 14 '12 at 0:27
    
Thanks so much for your answer! This clears up a lot of confusion. Yes I think we need open $n$ - disks. Now in the second case, I don't think I have been able to go anywhere. In Hatcher I believe he just says that $U$ deformation retracts onto $A$ via handwaving. Can you expand on that a bit more please? –  fpqc Aug 14 '12 at 0:30
    
@Mila I am having trouble seeing the deformation retract even in the case where $U$ is $X$ minus a point. –  fpqc Aug 14 '12 at 0:42
    
@BenjaLim I added a brief explanation of the deformation retraction. I didn't go into too much detail because a) in my experience no one goes into much detail in algebraic topology and b) when you do go into detail, you should do it yourself, because you'll get lost in other people's formulas otherwise. –  Miha Habič Aug 14 '12 at 0:56
    
Thanks so much for your time in explaining to me these details. I will accept your answer. –  fpqc Aug 14 '12 at 1:04
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