Let $f:X\rightarrow Y$ a continuous and surjective function.
(1) For every $y\in Y$ and for every $U$ open in $X$, if $f^{-1}(y)\subseteq U$ then $y\in int(f(U))$.
(2) For all $B\subseteq Y$ the map $f|_{f^{-1}(B)}:f^{-1}(B)\rightarrow B$ is a quotient map.
Prove that (1) $\Rightarrow$ (2).
We know that the restriction $f|_{f^{-1}(B)}$ is continuous because $f$ is continuous, then the topology $\tau_{B}$ of $B$ is contained in the final topology $\tau_{f|_{f^{-1}(B)}}$ generated by $f|_{f^{-1}(B)}$. I'm stuck with the other inclusion, any help will be appreciate.
Thank you.
A direct way to show that a function $g: A \mapsto B$ is a quotient map is to check that (1) $g$ is surjective, and (2) a set $V \subseteq B$ is open in $B$ if and only if its preimage set $g^{-1} (V)$ is open in $A$.
For notational simplicity, let's call $g = f \big|_{A}: A \mapsto B$, where $A = f^{-1} (B) \subseteq X$ and $B \subseteq Y$.
(1) Checking surjectivity of $g$ is easy: $f$ is surjective, which means for any $y \in Y$, there exists $x \in X$ such that $f (x) = y$. Take any element in the codomain of $g$, say $b$. Since $B \subseteq Y$ and $b \in B$, then $b \in Y$, so that there exists $a \in X$ such that $f (a) = b$. But $f (a) = b \in B \implies a \in A$, as desired.
(2) Suppose first that $V \subseteq B$ is open. Because $f$ is continuous, and the restriction of the domain of a continuous function is continuous, then $g$ is continuous (make sure you understand why). Therefore, $g^{-1} (V)$ is open in $A$.
Now suppose that $V \subseteq B$ is such that $g^{-1} (V) \subseteq A$ is open (in $A$). By definition, there exists $C$ open in $X$ such that $g^{-1} (V) = A \cap C$. Note also that $g^{-1} (V) = f^{-1} (V)$ (see below). It is easy to see that $f^{-1} (y) \subseteq C$, since if we have $x \in f^{-1} (y)$, then $f (x) = y \in V \implies x \in f^{-1} (V) \subseteq C$, so $x \in C$. Therefore, by hypothesis, $y \in {\rm Int} C$. By surjectivity of $f$, $V = f (f^{-1} (V)) = f (A \cap C) = B \cap f (C)$. The set $B \cap {\rm Int} C$ is at once open in $B$, contains the point $y$, and is contained in $V = B \cap C$ (since ${\rm Int} C \subseteq C$ is indeed true for any set). Since $y$ was arbitrary, $V$ is open.
Proof that $g^{-1} (V) = f^{-1} (V)$ for $V \subseteq B$
We have (1) $g$ is a restriction of $f$, showing that $g^{-1} (V) \subseteq f^{-1} (V)$, and (2) for any $x \in f^{-1} (V)$, we have $x \in f^{-1} (B)$ so that $g (x) = f(x) \in V$ and thus $x \in g^{-1} (V)$, showing that $f^{-1} (V) \subseteq g^{-1} (V)$. Take an arbitrary $y \in V$ (if such an element does not exist, $V = \emptyset$ and is clearly open). Let $x \in f^{-1} (y)$ be arbitrary. Then $f (x) = y \in V$ so that $x \in f^{-1} (V) = g^{-1} (V)$. Since $x$ was arbitrary, $f^{-1} (y) \subseteq g^{-1} (V)$.