Showing that a set is open in the strong topology induced by a family of functions

Question

Let $Y$ be an arbitrary set. There is a natural way in which a family of maps $f_i: X_i → Y$ (for $i ∈ I$) from topological spaces $X_i$ to $Y$ induces a topology on $Y$. Namely, the strong topology induced by the family {$f_i: i ∈ I$} is the finest topology on $Y$ with respect to which all the maps $f_i$ are continuous. The strong topology generalizes the quotient topology and can be thought of as a “dual” concept to the weak topology

Show that a set $V ⊆ Y$ is open in the strong topology induced by $f_i: X_i → Y$ (for $i ∈ I$), if and only if for each $i ∈ I$, the set $f^{−1}_i(V)$ is open in $X_i$.

My attempt:

=>: $f_i$ is continuous, then for every open $V⊆Y$, we have $f_i^{-1}(V)$ is open in $X_i$ for every $i \in I$ (by definition of continuity).

<=: $V = f_i(f_i^{-1})(V) = f_i^{-1}(f_i(V))$. But $f_i(V)⊆Y$, and $f_i$ is continuous, then for every $f_i(V)⊆Y$ open, $f_i^{-1}(f_i(V))$ is open, which implies that $V⊆Y$ is open.

Is my attempt correct?

Answer 1

Hint

Denote $$\sigma^\prime = \{V \subseteq Y \mid f_i^{-1}(V) \in \tau_i \, , \forall i \in I\}$$ where $\tau_i$ is the topology of $X_i$.

Prove that $\sigma^\prime$ is a topology.

Now, the very definition of continuity implies that every topology $\sigma$ on $Y$ for which all maps $f_i : (X_i, \tau_i) \to (Y, \sigma)$ are continuous is contained in $\sigma$. This $\sigma^\prime$ is the finest topology on $Y$ making all $f_i$ continuous, i.e. is the strong topology.