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I'm having a difficult time fully grasping what is happening in an adjunction space. Let me start by writing out the definition:

Let X, Y be topological spaces. $A$ is a closed subspace of $Y$, $f:A\rightarrow X$ is a continuous map. Let ~ be the equivalence relation on the disjoint union $X\bigsqcup Y$ generated by $a$~$f(a) \forall a \in A$, and denote the resulting quotient space by \begin{equation} X\bigcup_f Y=(X\bigsqcup Y)/\sim \end{equation} This is called the adjunction space and is said to be formed by attaching Y to X along $f$.

Ok, so I am trying to prove a proposition:

Let $ X\bigcup_f Y$ be an adjunction space and let $q:X\bigsqcup Y \rightarrow X\bigcup_f Y$ be the associated quotient map. Then the restriction of $q$ to $X$ is a topological embedding whose image set $q(x)$ is a closed subspace of $X\bigcup_f Y$.

The proof is:

We begin by showing $q|_X$ is a closed map, suppose $B$ is a closed subset of $X$, to show that $q(B)$ is closed in the quotient space, we need to show $q^{-1}(q(B))$ is closed in $X\bigsqcup Y$, which is equivalent to showing that its intersections with $X$ and $Y$ are closed in $X$ and $Y$ resp. (ALL OK SO FAR). From the form of the equivalent relation, it follows that \begin{equation} q^{-1}(q(B))\bigcap X=B\quad (*)\end{equation} which is closed in $X$ by assumption, and $q^{-1}(q(B))\bigcap Y=f^{-1}(B)$ which is closed in $A$ by continuity of $f$ and thus is closed in $Y$ because $A$ is closed in $Y$.

(*) is what I don't understand. It is a really simple statement, right? But I think it's my lack of total understanding of that is actually happening in the adjunct space that makes me unable to see why $q^{-1}(q(B))\bigcap X=B$. Is $q$ injective? It seems like $q$ must be atleast injective for this statement to hold, but I don't know how to show this.

Thanks for the help!

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    You have that $q$ is clearly surjective -- it is a quotient mapping. While $q$ will typically not be injective, $q|_X$ will be. Why? – 2017-02-06
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    The question you asked essentially is my struggle. Is $q|_X$ injective since $B\subseteq X$? – 2017-02-06
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    Actually no that doesn't make sense what I said. $q|X$ is a quotient map so its job is to identify points with each other. We are identifying points in $X\bigsqcup Y$ to their images in $X\bigcup_f Y$. So why does this guarantee that no element of $X$ gets mapped to the same element in $q(X)$? – 2017-02-06
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    Can you make an example where $q|_X$ is not injective? Where do your attempts fail? – 2017-02-06
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    I can kindddd of see whats happening but still need a little help. So basically this is an implication of the way we defined our attaching function, $f$, right? If we take an element w of $X\bigsqcup Y$, w is either in X or in Y, so we are just interested in $w\in X$. But $w=f(a)$ for some $a\in A$. And there was already an equivalence relation defined here, so in some way this "cleans up" any duplicate mappings. Can you help me word this in a better way? – 2017-02-06
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    Take $x \in X \subseteq X \sqcup Y$. What is $x$ equivalent to? It is equivalent to itself, and to each $a \in A$ with $f(a) = x$. Is it equivalent to anything else? Why or why not? – 2017-02-06
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    I'm not sure what is happening to the points of $Y\A$. This is the only place I could imagine something weird happening where $x\in X$ might be equivalent to something here? $f$ is not defined on $Y\A$ though so I dont see how we could make any equivalence – 2017-02-06
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    Indeed you can't. Points in $Y \setminus A$ are equivalent only to themselves. – 2017-02-06
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    Then I can't see how $x$ could be equivalent to anything else – 2017-02-06
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    Ahhh I see now. If q(c)=q(c') for c in X then both values are in the same equivalence class which could only be mapped to by one value of X which is why q is injective on X ! – 2017-02-06

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