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If $X,Y$ are CW complexes and $f$ a cellular map from $X$ to $Y$, is it true that $M_f$ is a CW complex?

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up vote 2 down vote accepted

Yes this is true. The proof can be found on Akhil Mathew's blog under 'The Loose End' at the bottom of the linked page.

To repeat what Akhil has on that page (If he reads this I hope he doesn't mind!)

First, it is a general and well-known fact that the product of two CW complexes is a CW complex. So $X \times I$ is a CW complex. And, from the construction, $X = X \times \left\{0\right\}$ is a subcomplex.

Let $X$ be a CW complex and $A \subset X$ a subcomplex. Suppose $f: A \rightarrow Y$ is a cellular map (for $Y$ another CW complex). Then the space $X \cup_f Y$ is a CW complex.

Proof: We take as the cells the cells of $X$ not in $A$ and the cells of $Y$. This is kosher, because $X-A$ is the union of various cells, $A$ being a subcomplex. The reason we have to require $f: A \rightarrow Y$ to be cellular, though, is that the boundary of each $n$-cell has to be contained in a union of $n-1$-cells.

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