Adjunction space quotient properties. Let $X\cup_f Y$ be an adjunction space. Let $q:X\coprod Y \rightarrow X\cup_f Y$ be the associated quotient map,
where $\sim$ is generated by $a\sim f(a)$ for all $a$.
Show that $q$ is injective.
My attempt:
Case 1: Let $(x_1,0),(x_2,0)$ ( I will write it as $x_1,x_2$) be in $X$. So $q(x_1)=q(x_2)\implies [x_1]=[x_2]$. Hence, $x_1\sim x_2$. 
This means that there exists $a_1,a_2....,a_n$ in $X\coprod Y$ such that
$a_1=x_1$ and $a_n=x_2$ and for each $i=1,2,...,n-1$.
$a_iRa_{i+1}$ or $a_{i+1}Ra_i$ or $a_i=a_{i+1}$ $(*)$
Clearly, the first and second cases can't hold. Hence, $x_1=x_2$.
Case 2: Suppose $x\in X$ and $y\in Y\backslash A$, then clearly $x\neq y$. I must show that $[x]$ and $[y]$ are disjoint. This is clearly the case, since it contradicts all of the three cases in $*$.
Case 3: Suppose $x\in X$  and $y\in A$. Then clearly $x\neq y$.
The two other cases follows similarly.
Is my attempt correct?
 A: $q$ is not necessarily injective on all of $X\sqcup Y$. In fact by definition $q(a) = q(f(a))$ for all $a \in A$, so $q$ is globally injective iff $A$ is empty. Your argument in case $3$ is not valid.
I think you have the right idea for Case 1 but I would get rid of the "clearly"s as follows. Let's say a "trivial relation" is one of the form $p\sim p$ and an "atomic relation" is one of the form $a\sim f(a)$ for $a\in A$; then every relation $p\sim p'$ which isn't trivial can be written as a sequence of atomic relations. Suppose now that $x\sim x'$ for $x \neq x' \in X$. Since there are no atomic relations between elements of $X$ or between elements of $Y$, there must be a sequence 
$$x = x_1\sim a_1 \sim x_2 \sim \dots a_n \sim x_n = x'$$
alternating between elements of $X$ and $A$. But for $1\leq i < n$ we would have $x_i \sim a_i \sim x_{i+1}$ and so $x_i = f(a_i) = x_{i+1}$, where we're implicitly using the fact that $f$ is well-defined. It follows that we must have $x = x'$. (I think this argument is basically the same as what you were going for, this is just the way it makes sense to me.)
By a similar argument you can prove that $q$ is injective on $Y$ iff $f$ is injective, since the function $f^{-1}\colon im(f) \to A$ would be well-defined. (But still $q$ cannot be injective on all of $X\sqcup Y$ unless $A$ is empty.)
A: Maybe you're aiming for this well-known fact: if $f: A \to Y$ and $A$ is closed, then $q$ restricted to $X\setminus A$ is an open embedding. 
