Let $X$, $Y$, $Z$ be topological spaces homeomorphic to CW complexes. And let $f:X\to Y$, $g:Y\to Z$ be cellular maps.


My question is "Is the composition $g \circ f$ cellular map?".


If $Y$ admits two different CW decomposition, I think this question becomes a little hard problem. How can it proved?



Is the composition of cellular maps cellular?
A CW complex $X$ includes the data of the skeleton filtration $X\_n$. Recall that, for $X$ and $Y$ CW complexes, a continuous map $f:X\rightarrow Y$ is called cellular if $f(X\_n) \subseteq Y\_n$ for all n. In this sense, the composition of two cellular maps is again cellular.


Here is a reformulation of the question: Let $X$, $Y$, $Y'$ and $Z$ be CW complexes such that the underlying topological spaces of $Y$ and $Y'$ are equal. If $f:X \rightarrow Y$ and $G: Y' \rightarrow Z$ are cellular maps, is the continuous map $g \circ f$ is cellular?


Answer: No. We can take the underlying topological spaces to be the interval $[0,3]$ for each of $X,Y,Y', Z$. For $f$ and $g$ we use the identity map. The identity map is cellular if and only if the 0-skeleton of the source is contained in the 0-skeleton of the target. On an interval we can specify CW structures by a finite subset of the interior. The 0-skeleton is then that subset together with the two endpoints.
For example, take $X\_0 = \{0,1,3\}$, $Y\_0 = \{0,1,2,3\}$, $Y'\_0 = \{0,2,3\}$, and $Z = Y'$ (equality of CW complexes). The composition $g\circ f$ is again the identity map on underlying topological spaces and it is not cellular because $X\_0$ is not a subset of $Z\_0$. 


Note that this answer should not be surprising because there is no condition on relating the CW structures on $Y$ and $Y'$. Another way of expressing this is that $f$ and $g$ are not composable morphisms in the category of CW complexes and cellular maps. 


