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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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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 $X\times Y$, that is the topology coherent with the compact subsets of that space, so $X\times_k Y$ is the product in the category of $k$-spaces. By $X\times_{CW} Y$ we will denote the CW structure on the product.

Since every CW complex is a $k$-space, being a quotient of a topological sum of balls, which are compact Hausdorff spaces, so is $X \times_{CW} Y$. The projections from $X \times_{CW} Y$ to $X$ and to $Y$ induce the identity map to $X\times Y$, and thus a continuous identity map $i: X \times_{CW} Y \to X\times_k Y$, by universality of the $k$-ification.

In order to show that the inverse $j: X \times_k Y \to X \times_{CW} Y$ is continuous, we show that $j|_K$ is continuous on every compact subset $K$ of $X\times Y$. Since the both projections of $K$ are compact, they are contained in finite subcomplexes $C_X \subset X$ and $C_Y\subset Y$. Their product $C_X\times C_Y$ is compact Hausdorff and thus a $k$-space, and it is a subspace of $X\times_k Y$ containing $K$. Since $C_X \times_{CW} C_Y \to C_X \times C_Y$ is a continuous map from a compact to a Hausdorff space, it is a homeomorphism. That means $j|_K$ embeds $K$ into $C_X\times_{CW} C_Y$ and thus into $X \times_{CW} Y$.

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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