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Problem of the Week #274 - Jul 17, 2018

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Let $X$ and $Y$ be normal topological spaces. Suppose $A$ is a closed subset of $X$ and $f : A \to Y$ is a continuous map. Prove that the adjunction space $X \cup_f Y$ is normal.
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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here's a hint: Consider the Tietze extension property.
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
No one answered this week’s problem. You can read my solution below.

It suffices to show that $X \cup_f Y$ has the Tietze extension property. Let $B$ be a closed subset of $X \cup_f Y$; let $p : X \cup Y \to X\cup_f Y$ be the projection map. Given a continuous map $g : B \to \Bbb R$, the restriction of $g$ to the closed set $p^{-1}\cap B$, $g_B: p^{-1}(B) \cap Y \to \Bbb R$, has a continuous extension $\phi : Y \to \Bbb R$ by normality of $Y$. The composition $g_B\circ f : A \to \Bbb R$ agrees with $\phi$ on the intersection $A\cap [p^{-1}(B) \cap X]$, so there is a natural extension $\phi : A \cup [p^{-1}(B) \cap X] \to \Bbb R$. Normality of $X$ gives a continuous extension of $\phi$, $\Phi : X \to \Bbb R$. Now $\Phi(a) = g_B(f(a))$ for all $a\in A$, so by the universal property of quotients $\Phi$ and $g_B$ induce a unique continuous map $G : X\cup_f Y \to \Bbb R$. This map extends $g$.
 
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