Let $(X,E)$ be a CW complex, and $Y$ be any space. Prove that a function $f:X\to Y$ is continuous if and only if $f\Phi_{e}$ is continuous for all $e\in E$.

The map $\Phi_{e}$ is the characteristic map for the cell $e$, i.e. $\Phi_{e}:(D^{n},S^{n-1})\to (e\cup X^{(n-1)},X^{(n-1)})$.

Necessity is obvious, so the sufficiency is the problem here. I tried to solve this problem by showing that restriction of $f$ to each closure $\bar{e}$ is continuous and use the weak topology of $X$. It is easy to prove that Restriction of $f$ to each 'open' cell $e$ is continuous since characteristic maps are relative homeomorphisms. But I can't figure out a way to extend this result to whole of $\bar{e}$. Can anyone help me with problem? Please enlighten me.

  • $\begingroup$ What is $\Phi_e$? $\endgroup$ – user99914 Aug 28 '16 at 6:37
  • $\begingroup$ The characteristic map for the cell $e$. I will add it in the question. $\endgroup$ – Dilemian Aug 28 '16 at 6:46
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    $\begingroup$ A CW complex is the colimit of its cells, endowed with its weak topology. This is just the universal property of such a colimit —most of the information making the space a CW complex is irrelevant. $\endgroup$ – Mariano Suárez-Álvarez Aug 30 '16 at 1:24

The abstract argument, as already mentioned in the comments, is the following:

Any CW complex is the colimit of its cells. In general, if you have a diagram $D$ of topological spaces, then a map $$\operatorname{colim} D\longrightarrow Y$$ is continuous if, and only if it is continuous on the restriction to every space in the diagram.

We can also dirty our hands and go for a direct proof:

One direction is obvious: if $f$ is continuous, then $f\Phi_e$ will also be continuous for any $e$.

Conversely, assume $f\Phi_e$ is continuous for all $e$. Let $U\subseteq Y$ be open. We have to show that $f^{-1}(U)$ is open. But this is true if, and only if $f^{-1}(U)$ intersects every cell $e$ in an open subset. Now,its intersection with $e$ is given precisely by $\Phi_e^{-1}(f^{-1}(U)) = (f\Phi_e)^{-1}(U)$, which is open by continuity of $f\Phi_e$. Therefore, $f^{-1}(U)$ is open. This implies that $f$ is continuous.

  • $\begingroup$ $X$ has the weak topology from the family $\{\bar{e};e\in E\}$, so a subset of $X$ is open iff its intersection with each $\bar{e}$ (not $e$) is open in $\bar{e}$. Does the same hold for the family $\{e\}$?? $\endgroup$ – Dilemian Sep 6 '16 at 12:48

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