Topology induced by maps and continuity of $g$ if and only if $f_i \circ g$ is continuous for all $i$

Question

Let $X$ be a set and $(f_i)_{i \in I}$ a family of maps $f_i : X \to Y_i$, where $(Y_i, \tau_i’)$ is a topological space for each $i$. Show that the topology $\tau$ incuded by the family has the following property. Let $(Z, \tau’’)$ be a space and $g:Z \to X$, then $g$ is continuous if and only if $f_i \circ g$ is continuous for all $i$. Show also that the induced topology $\tau$ is the only topology on $X$ with the mentioned property.

Suppose that $g$ is continuous, then for any $O$ open in $X$ we have that $g^{-1}(O)$ is open in $Z$. But any $O$ open in $X$ is of form $f^{-1}_{i_1}(V_{i_1}) \cap \dots \cap f^{-1}_{i_m}(V_{i_m})$ so $g^{-1}(O)= g^{-1}\left( f^{-1}_{i_1}(V_{i_1}) \cap \dots \cap f^{-1}_{i_m}(V_{i_m}) \right) = \bigcap_{k=1}^m (f_{i_k}^{-1} \circ g^{-1})(V_{i_k})$ for any $i \in I$ so $f_i \circ g$ is continuous?

Answer 1

Almost: it's not true that any open $O$ is of that form, but basic open sets are of that form.

Show two directions: if $g$ is continuous, all $f_i \circ g$ are continuous as compositions of two continuous maps. You have to note this separately.

Now assume all $f_i \circ g$ are continuous and we want to show $g$ is. So it suffices to show that a basic open set $O=f^{-1}_{i_1}(V_{i_1}) \cap \dots \cap f^{-1}_{i_m}(V_{i_m})$ has open inverse image (in fact, a subbasic $O=f_i^{-1}[V]$ is all you need to consider, but I'll follow your preference for using the base instead) then indeed

$$g^{-1}[O] = \bigcap_{k=1}^m g^{-1}[f^{-1}_{i_k}[V_{i_k}] = \bigcap_{k=1}^m (f_i \circ g)^{-1}[V_{i_k}]$$

(the last step is needed) and $g^{-1}[O]$ is a finite intersection of open sets because we assumed all $f_i \circ g$ are continuous (!). Note the direction in the proof. So $g$ is indeed continuous.

OK, let $\tau'$ be any topology on $X$ that obeys this same map-continuity property.

Define $g(x)=x$ (the identity) from $(X,\tau) \to (X,\tau')$. Is $g$ continuous? Check using the property for $\tau$: $f_i \circ g: (X,\tau) \to (Y_i, \tau'_i)$ is just the map $f_i$ itself (on $\tau$) and so is continuous. It follows that $g$ is continuous indeed. It follows that $\tau' \subseteq \tau$. Using the identity in the other direction and the property of $\tau'$ we symmetrically get $\tau \subseteq \tau'$ and so we have equality and unicity.

For more on these initial topologies on $X$ I refer to my post here. It also contains this proof. It's quite long and technical so for abstraction lovers only.