This is the theorem that I am trying to prove.
Theorem 1.3.6, Basic Algebraic Topology:
Let $q: X \to Y$ be a quotient map and $A \subseteq X$. Then the restriction map $q_A: A \to B=:q[A]$ is a quotient map if the following conditions are satisfied:
- $B$ is the intersection of an open and a closed set in $Y$.
- For $C \subseteq B$, $q_A^{-1}[C]$ is closed in $A$ (resp. open in $A$) iff $q^{-1}[C]$ is closed (resp. open) in $q^{-1}[B]$.
I was able to prove that $q_A$ is surjective and continuous. I am trying to show that
$C$ is open in $B$ if $q_A^{-1}(C)$ is open in $A$ for some $C \subset q(A)=B$
So assuming that $q_A^{-1}(C)$ is open in $A$ for some $C \subset B$, using the condition $(ii)$, I get $q^{-1}(C)$ is open in $q^{-1}(B)$.
Also, from condition $(i)$, $B=U \cap F$ for some $U$ open in $Y$ and $F$ closed in $Y$; This gives
$q^{-1}(B) = q^{-1}(U) \cap q^{-1} (F)$
i.e. $q^{-1} (B)$ is open in $q^{-1}(F)$ since $q^{-1} (U)$ is open in $X$
So, $q^{-1} (C)$ is open in $q^{-1} (F)$.
How do I proceed further ?