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I was reading the solution to this question:

If $\{q_\alpha: X_\alpha \to Y_\alpha\}$ is a family of quotient maps, then $q:\coprod_\alpha X_\alpha \to \coprod_\alpha Y_\alpha$ is a quotient map.

Specifically the part of ###Existence theorem for final topologies:### and I looked at the definition of $\mathcal{T_{f}}$

Here is it: $$\mathcal{T}_f = \{O \subseteq X: \forall i \in I: (f_i)^{-1}[O] \in \mathcal{T}_i \}$$

but it seems for me that it is not a topology as if $O_{1}$ and $O_{2}$ are in $\mathcal{T_{f}}$ then we can not proof that $O_{1} \bigcup O_{2}$ is in $\mathcal{T_{f}}$ as $f_{1}^{-1}[O_{1}] \bigcup f_{2}^{-1}[O_{2}] \neq $ anything as we have $f_{1}$ and $f_{2}$, am I correct ?if so, do we need to let this $\mathcal{T_{f}}$ a subbase?

Intuition
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1 Answers1

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This topology is defined very differently: it’s not a collection of inverse images of open sets ( those live in the domains of the $f_i$) but a set of subsets of the common codomain that obey the condition that for all $i$, the set $f_i^{-1}[O]$ is open in the given domain space $X_i$. You can see it as an intersection of topologies on $Y$, one to make each $f_i$ continuous. So we deal with inverse images of each $f_i$ separately.

E.g. to see intersection of two sets: let $O_1,O_2 \in \mathcal{T}_f$. To check that $O_1 \cap O_2 \in \mathcal{T}_f$ we can say: let $i \in I$ be arbitrary. Then $f_i^{-1}[O_1] \in \mathcal{T}_i$ and $f_i^{-1}[O_2] \in \mathcal{T}_i$. As $\mathcal{T}_i$ is a topology and $$f_i^{-1}[O_1 \cap O_2] = f_i^{-1}[O_1] \cap f_i^{-1}[O_2] \in \mathcal{T}_i$$ and as $i$ was arbitrary, $O_1 \cap O_2 \in \mathcal{T}_f$. Unions are similar.

In the common domain case, we’re taking a union of topologies, all consisting of inverse images under different $ f_i$ and unions of topologies aren’t topologies in general, so we use that union as a subbase to generate a topology to get around that issue.

Henno Brandsma
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