Elementary topology problem. Let $ ((Y_{\alpha},\tau_{\alpha}) \mid \alpha \in J) $ be a $ J $-indexed family of topological spaces and $ X $ any non-empty set. Let $ (f_{\alpha} \mid \alpha \in J) $ be a $ J $-indexed family of functions, where $ f_{\alpha}: Y_{\alpha} \to X $. What topology $ \tau $ can you put on $ X $ that will make all of the $ f_{\alpha} $’s continuous with respect to the $ \tau_{\alpha} $’s and $ \tau $?
Please help me with this. I think that I only need to put the indiscrete topology on $ X $, so that the only open sets in $ X $ are $ X $ itself and $ \varnothing $, and the inverse image of each of these under $ f_{\alpha} $ is open in $ Y_{\alpha} $.
 A: One can define a topology $ \tau $ on $ X $ as follows:

Declare a subset $ U $ of $ X $ to be $ \tau $-open if and only if $ {f_{\alpha}^{\leftarrow}}[U] \in \tau_{\alpha} $ for each $ \alpha \in J $.

Then $ \tau $ is the finest topology on $ X $ that makes $ f_{\alpha}: (Y_{\alpha},\tau_{\alpha}) \to (X,\tau) $ continuous for each $ \alpha \in J $.
A: First establish what subsets of $X$ can be open and which subsets cannot be open. We want the inverse of an open set to be open, so we should not let $S \subset X$ be open if there is at least one $\alpha$ for which $f_{\alpha}^{-1}(S) \notin \tau_{\alpha}$. Thus the only candidates for open sets are those whose inverse is open under every function in family.
Now, it may not be immediately obvious that the collection of all subsets which satisfy this latter property will form a topology. Perhaps we can take the union of such sets whose inverses are open, and then get a set whose inverse it not open? However, you can check the topology axioms yourself and see that the collection of all such sets is indeed a topology, so this is certainly the biggest topology that makes all the functions continuous, as the only sets we left out are those that explicitly ruin continuity for some function.
