Let $X$ be a CW-complex and $Y$ a subcomplex. Show $X/Y$ is a CW complex.

Question

As the title suggests I want to show that for $X$ be a CW-complex and $Y$ a subcomplex, it follows that $X/Y$ is a CW complex. Recall the following definitions

Definition: $X$ is a CW-complex if we have a sequence of subspaces $X_0\subset X_1 \subset \ldots \subset X$ such that

1. $X_0$ is discrete and each $n > 0$ the space $X_n$ is obtained by attaching $n$-cells to $X_{n-1}$

2. $X=\bigcup_nX_n$

3. $U\subset X$ is open $\iff$ $U_n=U\cap X_n \subset X_n$ is open for all $n$

.

Definition: $Y$ is a subcomplex of $X$ if $Y_n=Y\cap X_n$ and $Y_n\backslash Y_{n-1}$ is a union of open $n$-cells in $X_n \backslash X_{n-1}$.

This means we have to show:

1. $X/Y=\bigcup_n(X_n/Y_n)$

2. $U\subset X/Y$ is open $\iff U_n=U \cap (X_n/Y_n)\subset(X_n/Y_n)$ is open

Attempt: Let $U$ be open in $X/Y$, then $q^{-1}(U)$ is open in $X$. Since $X$ is a CW-complex, $q^{-1}(U)\cap X_n$ is open in $X_n$. But $V$ is open in $X_n/Y_n$ if and only if $q^{-1}(V)$ is open in $X_n$. This feels sketchy but I think $q^{-1}(U\cap(X_n/Y_n))=q^{-1}(U)\cap X_n$.

This doesn't seem correct as I have not used the fact that $Y$ is a subcomplex. I also do not know how to show the requirements.

Could you provide the details for this argument?

Answer 1

You use that $Y$ is a subcomplex to show that $X/Y= \bigcup_n X_n/Y_n$ (which you haven't made precise, and haven't proved yet); and to prove that $X_{n+1}/Y_{n+1}$ is obtained from $X_n/Y_n$ by adding $n+1$-cells (which you haven't proved either)

Now, to complete the argument, here's a sketch :

1- You have to make sense of $\bigcup_n X_n/Y_n$ : for that, try to define a map $X_n/Y_n\to X_{n+1}/Y_{n+1}$ which "looks like" the injection $X_n\to X_{n+1}$

2- Prove that $X/Y$ is the colimit of the $X_n/Y_n$ (using the above maps). Explicitly, that means prove that you have maps $X_n/Y_n\to X/Y$ that are compatible together (i.e. if you go from $X_n/Y_n$ to $X_{n+1}/Y_{n+1}$ then to $X/Y$, it's the same as going directly to $X/Y$), such that any point of $X/Y$ is in the image of some $X_n/Y_n$; such that (here it boils down to that, but watch out, if you encounter other colimits it will not always be the case) $X_n/Y_n\to X/Y$ is injective; and finally satisfying condition 3. as you tried to prove in your original question.

This amounts to proving that the family "image of $X_n/Y_n$ in $X/Y$" satisfies conditions 2. and 3. in your definition

3- Prove that the map $X_n/Y_n\to X_{n+1}/Y_{n+1}$ exhibits $X_{n+1}/Y_{n+1}$ as $X_n/Y_n$ to which we attached $n+1$-cells.

For this, use that $Y_{n+1}$ is $Y_n$ with cells attached, and $X_{n+1}$ is $X_n$ with the same cells + possibly other cells attached. This means we want to prove that $X_{n+1}/Y_{n+1}$ is $X_n/Y_n$ to which we attach these possible other cells.

If you know that "attaching cells" corresponds to pushout diagrams, you can proceed as follows :

you have two pushout diagrams $\require{AMScd}\begin{CD} S^n \times I_n @>>> Y_n \\ @VVV @VVV \\ D^{n+1}\times I_n @>>> Y_{n+1}\end{CD}$, $\begin{CD} S^n\times J_n @>>> X_n \\ @VVV @VVV \\ D^{n+1}\times J_n @>>> X_{n+1}\end{CD}$

with $I_n\subset J_n$ the sets of $n+1$-cells, and you have an inclusion of the first into the second, which produces a big cubical diagram; and then you just prove that "modding out" produces the following pushout diagram

$\begin{CD} S^n\times (J_n\setminus I_n) @>>> X_n/Y_n \\ @VVV @VVV \\ D^{n+1}\times (J_n\setminus I_n) @>>> X_{n+1}/Y_{n+1}\end{CD}$

(except perhaps for $n=0$, or if $Y$ is empty, but you can check those cases by hand) One way to do that is if you know some category theory, then colimits commute with colimits. If you don't know what that means, then you'll have to do it by hand, but it shouldn't be too complicated in this situation.