Quotient Topology, why is this set "saturated"?

Question

It says $[2,3]$ is saturated with respect to $q$, but not open in $Y$.

BUt it doesn't make sense to me because

$q(A) = q([0,1) \cup[2,3]) = [0,1) \cup [2-1,3-1] = [0,1) \cup [1,2] = [0,2] = Y$, so the image is open in $Y$ as it is $Y$.

Answer 1

Note that $p$ is almost 1-1, the only exception is $p(1) = 1 = p(2)$, this is where the two closed intervals of $X$ are "glued together", giving the result $[0,2]$, the image of $p$. $p$ induces an equivalence relation on $X$ that only identifies $1$ and $2$ and no other points.

A set $B \subset X$ is saturated under $p$, iff $1 \in A$ implies $ \in A$ and vice versa. Such a set satisfies the condition that if it contains a point from some class under that equivalence relation, it contains all members of that class. And the only non-trivial class is $\{1,2\}$. Alternatively, a saturated set is of the form $p^{-1}[C]$ for some $C \subset Y$. Simple set theory shows that this is the same notion. But the classes perspective shows why it's called "saturated",more clearly I think.

Now, $q$ is made to be injective, as we remove $1$ from $X$. So the equivalence relation induced by $q$ is the trivial one (only identify a point with itself), an so all subsets of $A$ are saturated. So the only way this could be a quotient map is to be a homeomorphism and this $q$ is not ($A$ is disconnected, $Y$ is connected, e.g.). Or as your text states it: $q^{-1}[[2,3]]$ is saturated (all sets are under $q$) and open, but $[2,3]$ is not open.