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Let $X$ be a topological space and $\mathcal{C}_1, \mathcal{C}_2$ be partitions of $X$ such that $(X, \mathcal{C}_1)$ and $(X, \mathcal{C}_2)$ are both CW complexes. Does there always exist a partition $\mathcal{C}$ of $X$ such that $(X, \mathcal{C})$ is a CW complex and $\mathcal{C}$ refines both $\mathcal{C}_1$ and $\mathcal{C}_2$? Refinement of $\mathcal{C}_1$ means that for each $C \in \mathcal{C}$ there exists $C_1 \in \mathcal{C}_1$ such that $C \subset C_1$.

If not, does the result hold if we assume finite CW complexes? Normality? Regularity?

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No, this can fail rather horribly. For instance, let $X=[0,1]^2$ and consider two triangulations of $X$: one which splits it into two triangles by a straight diagonal line, and one which splits it by an oscillating path that goes back and forth across the diagonal line infinitely many times. It is clear that any common refinement would need to have all the intersection points of these two paths as vertices, and so would have infinitely many cells. But since $X$ is compact, it does not admit any CW complex structure with infinitely many cells.

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  • $\begingroup$ Thanks. Along these lines concrete characteristic functions for the wobbly cells: $X = [-1, +1]^2$, $f_1 : (-1, +1) \to X$ s.t. $f_1(x) = (x, (1/4) \sin(\frac{\pi/2}{1 - |x|}) (1 - |x|)$, $f_2 : (-1, +1)^2 \to X$ s.t. $f_2(x, y) = (x, (1 - y)f_1(x) + y)$. $\endgroup$ – kaba Jul 2 at 22:18

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