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I am currently reading through May's "Algebraic Topology" and in the chapter on CW-complexes he shows that a product of CW-complexes is again a CW-complex, because one can define product cells using the canonical homeomorphism $(D^{n}, S^{n-1}) \simeq (D^{p} \times D^{q}, D^{p} \times S^{q-1} \cup S^{p-1} \times D^{q})$.

However, I remember hearing that a product of CW-complexes need not be again a CW-complex, ie. the topology on it is not the correct one. Here, May is working in the category of compactly generated spaces and the topology on this product can in general be finer than the usual product topology.

Is it easy to see that is this case the topology on the product of CW-complexes is the right one?

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Well, I don't know whether the fact you asked is easy or not (to you), but there is a reference for you to go over : (Hatcher, Algebraic topology), Theorem A.6. – cjackal Dec 1 '12 at 16:08
I agree that the reference touches the subject I mention, but it gives sufficient conditions for the product in category of all topological spaces to be "the right one". My question is whether the compactly generated topology is always the right one. – Matthew Dec 1 '12 at 17:56
Theorem A.6 says that if X,Y are CW complex (so objects in the category of compactly generated spaces) then the product of X & Y in this category also has the CW structure. I think this is exactly what you want. – cjackal Dec 1 '12 at 20:13

Let $X$ and $Y$ be CW complexes. Let $X\times Y$ be the usual product (in $\mathbf{Top}$, the category of all spaces), and let $X\times_k Y$ be the $k$-ification of $X\times Y$, so $X\times_k Y$ is the product in $\mathscr U$, the category of $k$-spaces. By $X\times_c Y$ we will denote the CW structure on the product.

So we want to show that the identity $i:X\times_k Y\to X\times_c Y$ is a homeomorphism.

  • Let $j$ denote the inverse function to $i$. Since the CW complex $X\times_c Y$ is a quotient of a topological sum of balls $D_{\alpha,\beta}^{m+n}\cong D_α^m\times D_β^n$, each with a characteristic map $\Phi_{α,β}:D_α\times D_β\to X\times_c Y$, it suffices to show that $j\circ\Phi_α$ is continuous for each $α$, i.e. that $\Phi_α\times\Phi_β$ is continuous to $X\times_k Y$. But since it is a test map to $X\times Y$ (a map from a compact Hausdorff space to $X\times Y$) and the $k$-product has the final topology with respect to all these test maps, it is also a map to the $k$-product. Hence $j$ is continuous.
  • To show that $i$ is continuous, we show that $i\circ t$ is continuous for any test map $t:K\to X\times Y$: Note that $t$ has coordinates $t_X, t_Y$, and each has a compact image which is thus contained in a finite subcomplex $C_X$ or $C_Y$ of $X$ or $Y$, respectively. Now the $k$-product $C_X\times_k C_Y$ (which is a subspace of $X\times_k Y$) is homeomorphic to the complex $C_X\times_c C_Y$: We have just shown that $C_X\times_c\times C_Y\to C_X\times_k C_Y$ is continuous, so, as a map from a compact to a Hausdorff space, it is a homeomorphism. That means $t$ is continuous as a map to $C_X\times_c C_Y$, thus $i\circ t$ is continuous.

I hope everything is correct, it is easy to lose track with all these different topologies flying around, so I urge you to double check what I did here.

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