The problem:
Let $\mathcal{F}$ be a family of closed subsets of a topological space $X$. Prove that if $Y$ is any space, then a function $f : X_w \to Y$ is continuous if and
only if $f|_F : F \to Y$ is continuous for all $F \in\mathcal{F}$. $X_w$ is the weak topology on $X$ relative to $\mathcal{F}$.
I want to know if my solution below is correct?
Let $V$ be a closed subset of $Y$. By the continuity of $f$, we have $f^{-1}(V)\subset U$ is closed where $U\subset X_w$ is closed and $U\cap F$ is closed in $F$ for all $F\in\mathcal{F}$. In particular, we have $f^{-1}(V)\cap F$ is closed in $F$ for all $F\in\mathcal{F}$. The restriction on $F$ gives the closed subset $f^{-1}|_F(V)\subset F$. Hence, function $f|_F$ is continuous.
Conversely, given that $f|_F$ is continuous, we have $f^{-1}|_F(V)$ is closed in $F$. For all $F\in\mathcal{F}$, $(U\cap f^{-1}|_F(V))\cup F$ is closed in $F$. $U\cap f^{-1}|_F(V)$ is closed. This implies there is a closed subset $V'$ of $V$ such that $f^{-1}(V')\subset U$ is closed. Thus, $f$ is continuous.
Best Answer
It's almost correct, but be more precise. The key observation is (for all $C \subseteq Y$ and all $F \subseteq X$).
$$(f\restriction_F)^{-1}[C] = f^{-1}[C] \cap F\tag{1}$$
which makes a simple connection between inverse images of the restricted map and those of $f$ and intersections with their domains. This is standard elementary set-theory.
So if $C \subseteq Y$ is closed and $f$ is continuous as a map from $X_w$, continuity of $f$ says that $f^{-1}[C]$ is closed in $X_w$, and $(1)$ read from left to right says that all restrictions of $f$ to $F \in \mathcal{F}$ indeed have a closed in $F$ inverse image of $C$, as $f^{-1}[C]\cap F$ is closed in $F$ by definition of $X_w$. So all $f\restriction_F$ are continuous by the closed-inverse image property.
OTOH if all restrictions are continuous and $C \subseteq Y$ is closed then $f^{-1}[C]$ is $X_w$ closed because $(1)$ tells us that any intersection with an $F \in \mathcal{F}$ is just the inverse image of $C$ of the restricted $f$ which is assumed to be closed by the continuity of $f\restriction_F$. So $f: X_w \to Y$ is continuous.