Is the adjunction space $X \cup_f Y$ normal for closed $A$ and continuous $f$?

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SUMMARY

The adjunction space $X \cup_f Y$ is proven to be normal when $X$ and $Y$ are normal topological spaces, $A$ is a closed subset of $X$, and $f: A \to Y$ is a continuous map. This conclusion is established by utilizing the Tietze extension property, which allows for the extension of continuous functions defined on closed sets. The discussion emphasizes the importance of these properties in confirming the normality of the adjunction space.

PREREQUISITES
  • Understanding of normal topological spaces
  • Familiarity with the Tietze extension property
  • Knowledge of continuous functions in topology
  • Basic concepts of adjunction spaces
NEXT STEPS
  • Study the Tietze extension theorem in detail
  • Explore examples of normal topological spaces
  • Research properties of adjunction spaces in topology
  • Examine continuous mappings and their implications in topology
USEFUL FOR

Mathematicians, particularly those specializing in topology, educators teaching advanced topology concepts, and students preparing for topology examinations will benefit from this discussion.

Euge
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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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Here's a hint: Consider the Tietze extension property.
 
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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