I am working on Problem 0.23 in Hatcher's Algebraic Topology text. It is as follows:
Show that a CW complex is contractible if it is the union of two contractible subcomplexes whose intersection is also contractible.
Suppose that $X$ is a CW complex such that $X = X_1 \cup X_2$, where $X_1$ and $X_2$ are contractible subcomplexes of $X$ such that $X_1 \cap X_2$ is contractible. Is it true that $(X,X_1 \cap X_2)$, $(X_1, X_1 \cap X_2)$ and $(X_2,X_1 \cap X_2)$ are all CW pairs? In other words, is is true that $X_1 \cap X_2$ is a subcomplex of $X$, $X_1 \cap X_2$ is a subcomplex of $X_1$, and $X_1 \cap X_2$ is a subcomplex of $X_2$?
The definition of a subcomplex of $X$ that I have is a closed subspace $A \subset X$ that is a union of cells in $X$. I think my trouble in answering this question is that I'm not sure how to interpret the intersection of two CW complexes.
All of the solutions to this exercise that I've seen (including this one) implicitly use the fact that all three of those are CW pairs, but I'm not quite seeing why. (Note: I highly doubt that we need the contractibility assumptions to answer this, but I included them anyways.)
Thanks!
Best Answer
This is true, even without the contractibility assumption (as you suspect). To see this, we need to merge the following two facts.
Fact 1: If $X$ is a CW complex and $X_1,X_2\subseteq X$ are subcomplexes, so is $X_1\cap X_2$.
Proof: Since $X_1,X_2\subseteq X$ are closed subspaces, so is $X_1\cap X_2$. Now, $X_1$ is a union of cells and cells are disjoint, so if a cell of $X$ meets $X_1$, it is contained in $X_1$. Analogously for $X_2$. Thus, if a cell of $X$ meets $X_1\cap X_2$, it meets both $X_1$ and $X_2$, hence is contained in both $X_1$ and $X_2$, i.e. contained in $X_1\cap X_2$. Since the cells cover $X$, it follows that each point of $X_1\cap X_2$ is contained in some cell of $X$, which necessarily meets $X_1\cap X_2$, hence is contained in $X_1\cap X_2$. Thus, $X_1\cap X_2$ is a union of cells.$\square$
Fact 2: If $X$ is a CW complex and $A\subseteq B\subseteq X$ are such that $A,B\subseteq X$ are subcomplexes, then $A\subseteq B$ is a subcomplex, when $B$ is given the CW structure inherited from $X$.
Proof: First, $A$ is closed in $B$ since $A$ is closed in $X$. Recall that the cells of the inherited CW structure of $B$ are just the cells of $X$ that are contained in $B$. Since $A\subseteq X$ is a subcomplex, $A$ is union of cells of $X$, but each of these cells is necessarily contained in $A\subseteq B$, so $A$ is also a union of cells of $B$.$\square$