# To show that product $Z=X×Y$ in the product topology is a CW complex

I would like to prove the following:

If $X,Y$ are CW complexes, and either $X$ or $Y$ is locally compact then the product $Z=X×Y$ in the product topology is a CW complex. (-see here)

In order to prove it, it needs to satisfy 3 conditions as in the definition of CW complex: I have proven the condition 1 and 2, they are quite straight forward and even hold without assumption that $Y$ being locally compact. I got stuck to prove the third condition, and I think it is where the locally compactness is used. So, suppose $A\cap \bar{e_\alpha}\times \bar{e_\beta}$ is closed in $\bar{e_\alpha}\times \bar{e_\beta},\forall\alpha,\beta$. How to show that $A$ is closed in $X\times Y$?

Any help is appreciated :)

I think this is related (or maybe follows from) the result by E. Michael that the product of a locally compact space and a $k$-space is again a $k$-space.

• Yes. There is such theorem in Hatcher that the product of a locally compact space and a k-space is again a k-space. Since $X$ is a CW complex, then it is directly a k-space. But now the result gives $X\times Y$ is a k-space, this is where I got stuck: how to show that it is CW? – Chen M Ling May 27 '16 at 10:38
• @ChenMLing you have to show that if we have a $k$-cover in every space, then the set of products from the $k$-covers is a $k$-cover in the product. I think this is done in the prood of this theorem, essentially. – Henno Brandsma May 27 '16 at 10:49

This follows from the standard result given in, for example, Topology and Groupoids:

$4.3.2$ Let $f : X \to Y$ be an identiﬁcation map and let $B$ be locally compact. Then
$$f × 1 : X × B \to Y × B$$ is an identiﬁcation map.

The definition of locally compact here is that each point has a base of compact neighbourhoods.

This inconvenient restriction led to the notion of convenient category of topological spaces. See also Section 5.9 of Topology and Groupoids.

• Thanks for your answer. But I do not learn about identification map so I think I wonder if I can prove it with the definition of CW complex. :) – Chen M Ling May 27 '16 at 10:41

I came about this post when I was reviewing my question. The proof by Eric Wolfsey can be modified:

We require two lemmas. The second result is a variant of general case of colimit topology.

Lemma 0: Let $Y$ be locally compact, $X \times Y$ have product topology, $Z^Y$ have comapct open topology. $f:X \times Y \rightarrow Z$ is continuous iff $\hat{f}:X \rightarrow Z^Y$ is continuous. [Tammo tom Dieck, pg37]

Lemma 1: Let $X$ be a Hausdorff space. $X = \bigcup_{e \in E} e$, disjoint union of cells. Let $\Phi_e:D_e^n \rightarrow X$ be the characteristic maps. Then $X$ has the colimit topology wrt closure of cells (i.e. $A \subseteq X$ closed iff $A \cap \bar{e}$ is closed for all $\bar{e}$) iff it satisfies the universal property: for all spaces $Z$, $f:X \rightarrow Z$ continuous iff $f \Phi_e:D^n_e \rightarrow Z$ is continuous.

Proof of lemma 1: First suppose that $X$ has the colimit topology. Then $$\bigsqcup \bar{e} \rightarrow X$$ is a quotient map. But we also have that $S:= \{\Phi^n_e:D^n_e\rightarrow \bar{e}\}$, with image restricted, is a quotient map, since it is a surjective closed map (we use the Hasudorff condition here). Thus, $$\bigsqcup D^n_e \rightarrow \bigsqcup \bar{e} \rightarrow X$$ is composition of two quotient maps. So the UP is satisfied wrt to $S$, using UP of quotient maps. $\Box$

Conversely, suppose the UP holds. Let $C$ be a set of $X$ such that $C \cap \bar{e}$ is closed in $\bar{e}$ for all $e$. Let $Z=\{0,1\}$, with $\{0\}, \{0,1\}$ as its open sets. Define $f(C)=\{1\}$. Then using Hausdorf property, we may deduce $\Phi^n_e(D^n_e)=\bar{e}$, giving $$(f\Phi_e)^{-1}(1)= \Phi_e^{-1}(C)=\Phi_e^{-1}(C \cap \bar{e})$$ is closed in $D^n_e$. Hence $f\Phi_e$ is continuous. Thus $f$ is continuous and $C$ is closed. $\Box$

Let $E,E'$ denote the cellular decomposition of $X,Y$ respectively.

1. $X \times Y$ has the cell decomposition $e \times e'$ for $e \in E$, $e' \in E'$. $X \times Y$ is unknown to have colimit topolgoy of these cells.
2. It suffices to show that $X \times Y$ satisfies the UP: for all spaces $Z$, $$f:X \rightarrow Z$$ is continuous iff $$f_{e,e'}:D_e \times D_{e'} \rightarrow Z$$ is continuous. Where $D_f$ are the closed $n$-balls, $n=\dim (f)$.

We simplify step 2. By lemma 0, as $D_e$ is locally compact, $f_{e,e'}:D_e \times D_{e'} \rightarrow Z$ is continuous iff $f^e_{e'}:D_{e'} \rightarrow Z^{D_e}$ is continuous for all $e'$. Since $Y$ has colimit topology, by equivlaence of lemma 1, $f^e_Y:Y \rightarrow Z$ is continuous iff $f^e_{e'}:D_{e'} \rightarrow Z^{D_e}$ is continuous for all $e' \in E'$. Then again, as $Y$ is locally compact, we can apply lemma 0, to deduce

1. It suffices to show the UP is satifised: $f:X \rightarrow Z^Y$ is continuous iff $f^e_Y:D_e \rightarrow Z^Y$ is continuous for all $e \in E$.

But this follows from lemma 1, as $X$ has colimit topology. $\Box$