# Munkres' Topology Problem

Can someone critic my proof?

Number 9 of page 112 of Munkres' Topology: Let ${\{A_{\alpha}\} }$ be a collection of subsets of $X$;let $X=\bigcup_{\alpha}A_{\alpha}$. Let $f:X\rightarrow Y$; suppose that $f_{|A_{\alpha}}$ is continuous for each $\alpha$.

Problem Show that if the collection ${\{A_{\alpha}\} }$ is finite and each set ${A_{\alpha} }$ is closed, then $f$ is continuous.

My attempt at proof : Let $C$ be closed in $Y$, then since: $$\bigcup_{\alpha}f^{-1}_{|A_{\alpha}}(C) = \bigcup_{\alpha}\left[f^{-1}(C) \bigcap A_{\alpha} \right]$$

$$\bigcup_{\alpha}f^{-1}_{|A_{\alpha}}(C) = f^{-1}(C) \bigcap X$$

$$\bigcup_{\alpha}f^{-1}_{|A_{\alpha}}(C) = f^{-1}(C)$$ Therefore $f^{-1}(C)$ is closed in X since it is a finite union of each closed sets $f^{-1}_{|A_{\alpha}}(C)$ (by hypothesis) in X. Hence $f$ is continuous.

I did not use the fact that each $A_{\alpha}$ is closed at all, this is why I think there must be something wrong with my proof. Anybody care to explain where I'm wrong?

• You did use that $A_{\alpha}$ is closed. Otherwise $(f\lvert_{A_{\alpha}})^{-1}(C)$ would generally not be closed in $X$. – Daniel Fischer Feb 28 '16 at 11:06
• @DanielFischer Hmmm. I am missing something. Why can the intersection not be closed while $A_\alpha$ aren't? – Rudy the Reindeer Feb 28 '16 at 11:07
• @DanielFischer I feel like a year ago I could do proofs like this with my eyes closed. Then I spent a year not doing any topology and now I can't even do the most basic thing. How depressing. Maybe my brain is broken. – Rudy the Reindeer Feb 28 '16 at 11:12
• @RudytheReindeer: Nah, your brain is not broken at all. It happens to everyone. terrytao.wordpress.com/career-advice/write-down-what-youve-done – Giuseppe Negro Feb 28 '16 at 11:19
• You could also view this as the pasting lemma + induction. – Justin Young Feb 28 '16 at 13:01

To elaborate on Daniel Fischer's comment: $f^{-1}_\alpha(C)$ is a priori closed in $A_\alpha$, not in $X$, since the map $f_\alpha$ (which was assumed to be continuous) has $A_\alpha$ as its domain, not $X$. You need to use that $A_\alpha$ is closed in $X$ to deduce that a closed subset of $A_\alpha$ in the subspace topology is actually closed in $X$.