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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?

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    $\begingroup$ You did use that $A_{\alpha}$ is closed. Otherwise $(f\lvert_{A_{\alpha}})^{-1}(C)$ would generally not be closed in $X$. $\endgroup$ – Daniel Fischer Feb 28 '16 at 11:06
  • $\begingroup$ @DanielFischer Hmmm. I am missing something. Why can the intersection not be closed while $A_\alpha$ aren't? $\endgroup$ – Rudy the Reindeer Feb 28 '16 at 11:07
  • $\begingroup$ @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. $\endgroup$ – Rudy the Reindeer Feb 28 '16 at 11:12
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    $\begingroup$ @RudytheReindeer: Nah, your brain is not broken at all. It happens to everyone. terrytao.wordpress.com/career-advice/write-down-what-youve-done $\endgroup$ – Giuseppe Negro Feb 28 '16 at 11:19
  • $\begingroup$ You could also view this as the pasting lemma + induction. $\endgroup$ – Justin Young Feb 28 '16 at 13:01
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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 $X$.

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  • $\begingroup$ Ahh, yes yes. I remember that theorem. Thanks, that means my proof works right? I just need to make that clear? $\endgroup$ – Kurome Feb 28 '16 at 11:10
  • $\begingroup$ Your proof looks okay to me otherwise, yes. $\endgroup$ – Stahl Feb 28 '16 at 11:11

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