Final Topology is Topology

Theorem
Let $X$ be a set.

Let $I$ be an indexing set.

Let $\left\langle {\left({Y_i, \tau_i}\right)} \right\rangle_{i \in I}$ be an $I$-indexed family of topological spaces.

Let $\left\langle {f_i: Y_i \to X} \right\rangle_{i \in I}$ be an $I$-indexed family of mappings.

Let $\tau$ be the final topology on $X$ with respect to $\left\langle{f_i}\right\rangle_{i \in I}$.

Then $\tau$ is a topology on $X$.

Proof
Define:
 * $\forall i \in I: \vartheta_i = \left\{{U \subseteq X: f_i^{-1} \left({U}\right) \in \tau_i}\right\} \subseteq \mathcal P \left({X}\right)$

Then, by the definition of intersection:
 * $\displaystyle \tau = \bigcap_{i \mathop \in I} \vartheta_i$

Since the intersection of topologies is a topology, it suffices to show, for all $i \in I$, that $\vartheta_i$ is a topology on $X$.

We now verify the three axioms for $\vartheta_i$ to be a topology on $X$.

Let $\mathcal A \subseteq \vartheta_i$.

Then, by Mapping Preimage of Union: General Result and by the definition of a topology, it follows that:
 * $\displaystyle f_i^{-1} \left({\bigcup \mathcal A}\right) = \bigcup_{U \mathop \in \mathcal A} f_i^{-1} \left({U}\right) \in \tau_i$

That is, $\displaystyle \bigcup \mathcal A \in \vartheta_i$.

Let $U, V \in \vartheta_i$.

Then, by Mapping Preimage of Intersection and by the definition of a topology, it follows that:
 * $f_i^{-1} \left({U \cap V}\right) = f_i^{-1} \left({U}\right) \cap f_i^{-1} \left({V}\right) \in \tau_i$

That is, $U \cap V \in \vartheta_i$.

By the definition of a topology, it follows that:
 * $f_i^{-1} \left({X}\right) = Y_i \in \tau_i$

That is, $X \in \vartheta_i$.