Properties of Adjunction Space 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!
 A: In my opinion, you may go back to the initial definition of the adjunction space and of the quotient map $q$.

*

*The equivalence relation (based on a continuous function $f$ on $A$) defining the adjunction space is stating the basic fact that "It identifies each point $x \in X$ with all of the points (if any) in $f^{-1}(x) \subset A$"

*Then, what does the quotient map $q$ do? You should bear in mind that the quotient map just maps all the points belonging to the same equivalence class to a single set(this set is the equivalence class). You can imagine this quotient map "glues" a $x$ and all points in $A$ which can map to $x$ (by $f$) together.

So, what does $q(B)$ do? It finds for all $x$'s in $B$ and "glues" those points which map to them, respectively.
For example, if the function $f$ is in the simplest form: $ a_1 \rightarrow b_1 $ and $ a_2 \rightarrow b_1 $ , leaving all other points unspecified. Then the result $q(B)$ shuold be like $\{ \{a_1,a_2,b_1\},\{b_2\}, \{b_3\}, \cdots \}$
Then, what does $q^{-1}q(B)$ look like? With the same example above, it just "unglue" those glued points, the  $q^{-1}q(B)$ looks like $ \{a_1,a_2,b_1,b_2, b_3, \cdots \}$
It follows that $q^{-1}q(B) \cap X = B$. It is also obvious that $q^{-1}q(B) \cap Y = f^{-1}(B)$.
Many times you come across problems, just go back to the definitions. You will find the answer.
If you need more clarification, let me know.
