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Given a space $X$ and a collection of subspaces $X_\alpha$ whose union is $X$, these subspaces generate a possibly finer topology on $X$ by defining a set $A\subset X$ to be open iff $A\cap X_\alpha$ is open in $X_\alpha$ for all $\alpha$. In case $\{X_\alpha\}$ is the collection of compact subsets of $X$, we write $X_c$ for this new compactly generated topology.

Given two CW complexes $X$ and $Y$, is $(X\times Y)_c\times I=(X\times Y\times I)_c$ established as true?

Suppose cells $e_X\subset X,e_Y\subset Y,e_I\subset I$, it's easy to know that $\overline{e_X}\times\overline{e_Y}\times\overline{e_I}$ has the same topology structure as subspace of $X\times Y\times I$ and of $(X\times Y)_c\times I$, thus $(X\times Y\times I)_c=((X\times Y)_c\times I)_c=(X\times Y)_c\times I$.

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From what I wrote at your other question, whenever $X$ is a $c$-space, then a map $X\to Y$ is also a map $X\to Y_c$, in particular $X_c\to X\to Y$ yields a map $X_c\to Y_c$ for any map $X\to Y$. If we apply this to the projections $X\times Y\to X$ and $X\times Y\to Y$, we get an isomorphism $(X\times Y)_c\to(X_c\times Y_c)_c$.

Now $I$ is a $c$-space since it is compact, so $I=I_c$. In order to show that for a $c$-space $X$, the product $X\times I$ is a $c$-space, you will need some facts:

  • A space $X$ is a $c$-space if and only if it can be expressed as a quotient of a disjoint union of compact spaces.
  • If $Y$ is locally compact, and $q:X\to Z$ is a quotient map, then $q\times\mathbf 1_Y:X\times Y\to Z\times Y$ is a quotient map.

With this you should be able to establish a homeomorphism $(X_c\times I)\to (X\times I)_c$. The fact that $X$ is a product of CW complexes is completely irrelevant.

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