Is the axiom of choice needed to show that the initial topology is a topology?

Question

Given any function $f:X \to Y$ and any topology $\tau_Y$ on $Y$, there is an induced topology $\tau_X$ on $X$ (called the initial topology) whose open sets are the inverse images of the open sets in $\tau_Y$ under $f$.

Now, the inverse image of a union is the union of the inverse images. But given a family of sets $(U_i)_{i \in I}$ in $\tau_X$, it seems that one needs to use the axiom of choice to choose for each $U_i$ an open set $V_i \in \tau_Y$ for which $f^{-1}(V_i)=U_i$. Once that is done, one could then show that the union of the $U_i$s is also in $\tau_X$.

But is the axiom of choice really needed?

I think that the answer is no, since if one does not know how to choose a $V_i$, then one could simply take the union of all the open sets in $\tau_Y$ that would work for $V_i$.

Answer 1

The general wisdom of choiceless proofs is "why choose at all?"

Instead, given $\{U_i\mid i\in I\}$ consider $\{V\in\tau_Y\mid\exists i, f^{-1}(V)=U_i\}$. Then this is a family of open sets, and since preimages play nice with the unions, its union is open in $\tau_Y$ and is the union of the $U_i$ in $\tau_X$.

Answer 2

For $U \in \tau_X$ let $$[U] = \{ V \in \tau_Y \mid f^{-1}(V) = U \}$$ and $$\phi(U) = \bigcup_{V \in [U]} V \in \tau_Y.$$ This gives a function $$\phi : \tau_X \to \tau_Y$$ which does not rely on any choice. We have $$U = \bigcup_{V \in [U]} f^{-1}(V) = f^{-1}(\bigcup_{V \in [U]} V) = f^{-1}(\phi(U))$$ and therefore $$\bigcup_{i \in I} U_i = \bigcup_{i \in I} f^{-1}(\phi(U_i)) = f^{-1}(\bigcup_{i \in I} \phi(U_i)) .$$ Since $\bigcup_{i \in I} \phi(U_i) \in \tau_Y$, we see that $\bigcup_{i \in I} U_i \in \tau_X$.