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

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 $$V_i$$.

• Try to write out the proof in your other way, if you think the answer is no. Sep 22 at 15:09
• You are correct, you can just use all choices at once. Sep 22 at 15:13
• There is a related problem. If we define a topology $\tau$ in the usual fashion, then the usual proof uses choice to show the following: "If $U\subseteq X$ has the property that, for each $x\in U,$ there is a $V\in \tau$ such that $x\in V\subseteq U,$ then $U\in\tau.$" The usual proof of this requires us to pick $V_x$ for each $x,$ which seems to require choice. Again, we can solve this by using all open $V$ which are contained in $U.$ Sep 22 at 15:25

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 $$\tau_X$$.
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 $$\bigcup_{i \in I} U_i \in \tau_X$$.