Let $(X, \mathscr T_X)$ and $(Y,\mathscr T_Y)$ be topological spaces.
GOAL: I wish to prove,
Let $J$ be a set and denote by $\mathscr F (J,Y)$ the set of all functions from $J$ to $Y$.
We equip it with the smallest topology containing $\{f \in \mathscr F (J,Y): f(j) \in V \}$ for every $V \in \mathscr T_Y$ and $j \in J$.
Show that a function $\Phi: X \to \mathscr F (J,Y)$ is continuous $\iff $ for each index $j\in J$ we have that $X\to Y$ given by $x \mapsto (\Phi(x))(j)$ is continuous.
Thoughts.
This indexing reminds a bit of the final topology, but it seems to be phrased a little bit differently (is this intuition correct?). Usually one direction of a proof like this is easy, and the other direction should be a bit more challenging. I am however struggling with understanding the opens of the space $\mathscr F (J, Y)$. Usually the smallest topology containing some subset is referred to as being generated as a subbase. In an earlier question we proved that it is sufficient to check the continuity requirement for elements of the subbase. I have proved the following lemma in the previous question:
Let $Y$ be a set and $S \subset \mathscr P (Y)$ a collection of subsets of $Y$. Equip $Y$ with the smallest topology $\mathscr T_Y$ on $Y$, containing $S$. Show that a function $f: X \to Y$ is continuous if and only if $f^{-1}(U)$ is open in $X$ for every $U \in S$.
It seems that I will be able to use this lemma, but I am a bit overwhelmed as where to start and in which direction to move. Can somebody help me with a rough outline (what is the easy direction, what do I wish to show, how do I use the properties?)
What I have so far (not a lot):
"$\implies$"
Suppose that $\Phi: X \to \mathscr F (J,Y)$ is continuous. Let $j\in J$. We look at $g_j: X \to Y$ defined by $g_j(x)=(\Phi(x))(j)$. Let $V$ be an open set in $Y$, so $V\in \mathscr T_Y$.
"$\impliedby$"
Assume that for each index $j \in J$ we have that the map $g_j:X \to Y$ defined by $g_j (x) =(\Phi(x)) (j)$ is continuous. Let $V$ be open in $Y$. Then we have $V \in \mathscr T_Y$. Now also let $j\in J$ be fixed. We then have that $g_j: X \to Y$ is continuous so $g_j^{-1}(V) \subset \mathscr F (J,Y)$ is open. …
Best Answer
HINT: For each $j\in J$ let $Y_j$ be a copy of the space $Y$; then $\mathscr{F}(J,Y)$ with this topology is just the product $\prod_{j\in J}Y_j$ with the product topology. The set
$$\{f\in\mathscr{F}(J,Y):f(j)\in V\}\tag{1}$$
is the same as $\prod_{i\in J}U_i$, where $U_j=V$, and $U_j=Y_j=Y$ for $j\in J\setminus\{i\}$, and those sets are the usual subbase for the product topology. Alternatively, if $\pi_j$ is the usual projection map from this product to the $j$-th factor, the set in $(1)$ is simply $\pi_j^{-1}[V]$.