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In Willard's General Topology, p66, section 9.14,

the author is trying to prove that $q(Y)$ is closed in $X +_{f} Y$.

$q$ is defined as the decomposition map of $X+Y$ onto $X +_{f} Y$ where f is a continuous function $f: A\subset X \rightarrow Y$.

$X +_{f} Y$ is the attaching of $X$ to $Y$ by $f$.

In the proof the author is stating:

Let $F$ be a closed subset of $Y$. Then $F$ is a closed subset of $X + Y$ and $F = q^{-1}[q(F)]$.

However, I don't see how this can hold. Is he confusing $q$ and $q|Y$ here? But even so I don't see the path.

almaus
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1 Answers1

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First of all Willard is trying to prove that $q|Y$ is a closed map, not that $q(Y)$ is a closed subset. The claim that $q|Y$ is closed is a lot stronger.

Let $F$ be a closed subset of $Y$. Then $F$ is a closed subset of $X + Y$ and $F = q^{-1}[q(F)]$.

Yes, Willard meant $q|Y$ here. This isn't true for $q$.

Denote $q':=q|Y$. The equality $F=q'^{-1}[q'(F)]$ follows because $q'$ is injective (see here). And it is injective because no two distinct points in $Y$ are related, as you can verify by careful analysis of $X+_f Y$ construction.

Side note: $X+_f Y$ is more often denoted by $X\cup_f Y$ and called the adjunction space.

freakish
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  • I'm still trying to understand the reasoning of the author: $q'$ is not anymore a quotient map (because not surjective), so why use that in combination with $F = q'^{-1}[q'(F)]$ in order to deduce first that $F$ is closed in $X +_f Y$? As $q(Y)$ is the codomain of $q'$ and that if we redefined it only on this codomain $q'': Y \subset X + Y \rightarrow q(Y) \subset X +_f Y$ it would become again a quotient map, then we obtain directly that $F$ is closed in $q(Y)$ isn't it? – almaus Oct 24 '19 at 16:10