Properties of Adjunction Spaces proof from Lee's Topological Manifolds

Question

I'm reading the proof to Proposition 3.77(b) in John M. Lee's Introduction to Topological Manifolds but I don't 100% follow the proof.

The proposition states: Let $X \cup_f Y$ be an adjunction space and let $q : X \coprod Y \rightarrow X \cup_f Y$ be the associated quotient map. Then, the restriction of $q$ to $Y \setminus A$ is a topological embedding whose image set $q(Y \setminus A)$ is an open subspace of $X \cup_f Y$.

The proof uses Proposition 3.62(d) from the same book -- if $q : X \rightarrow Y$ is a quotient map and $U \subseteq X$ is a saturated open/closed set, then the restriction of $q$ to $U$ is also a quotient map.

The proof is as follows:

• We note that $Y \setminus A$ is a saturated open subset of $X \coprod Y$, and so the restriction $q \rvert_{Y \setminus A}$ is also a quotient map.
• Further, since it is a bijection, the map is a homeomorphism. Thus, the map $q \rvert_{Y \setminus A}$ is a topological embedding.
• Finally, the image $q(Y \setminus A)$ is open in $X \cup_f Y$ by the definition of the quotient topology.

My doubts:

1. I think the reason it's a bijection is because $q$ is surjective by definition and injective because the equivalence relation does not identify and two points in $X \setminus A$ with each other (since only with points $a \in A$ are subject to the equivalence relation). Is this correct?

2. I don't quite know how to show that $Y \setminus A$ is saturated with respect to $q$. I think we have to show that $Y \setminus A = q^{-1}(q(Y \setminus A))$. Would this follow from the fact that none of the points in $X$ get identified with $X \setminus A$ by the map $q$?

Any hints/explanations would be appreciated!

Answer 1

$q|_{Y\setminus A}:Y\setminus A\to q(Y\setminus A)$ is surjective not because $q$ is generally a surjective map $X\sqcup Y\to X\cup_f Y$ but literally by definition of $q(Y\setminus A)$; the codomain here is defined to be the set of all points in the image of $q|_{Y\setminus A}$.

The only things that matter are checking that it is injective and that it is a quotient map onto its image. Because $A$ is closed in Lee's convention for adjunction spaces, $Y\setminus A$ is open. Is it saturated? Yes, for the reasons you say. $a\in A$ is identified with $f(a)\in X$ and then we take a transitive closure of the relation, but the final equivalence relation identifies $y\in Y\setminus A$ only with itself, $y\sim z\iff y=z$. This is also the reason why $q|_{Y\setminus A}$ injects; $q(y)=q(y')$ iff. $y\sim y'$ iff. (since $y,y'\notin A$) $y=y'$.

Let's really make sure. Suppose $q^{-1}(q(Y\setminus A))$ contains an element of $X\sqcup Y$ not in $Y\setminus A$, say some $a\in A$. Then $q(a)=q(y)$ for some $y\in Y\setminus A$ so $a\sim y$ so (as $y\in Y\setminus A$) $a=y$, but this is a contradiction. Now suppose it contains an element $x\in X$. $q(x)=q(y)$ for some $y\in Y\setminus A$ and again you conclude $x=y$ by definition of the equivalence relation, but $x\in X$ - $x\notin Y$ - so this is also a contraadiction. Therefore $q^{-1}q(Y\setminus A)\subseteq Y\setminus A$. Obviously this implies $q^{-1}q(Y\setminus A)=Y\setminus A$.

Both your doubts are ok.