Final Topology is Topology

Theorem
Let $X$ be a set.

Let $I$ be an indexing set.

Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.

Let $\family {f_i: Y_i \to X}_{i \mathop \in I}$ be an $I$-indexed family of mappings.

Let $\tau$ be the final topology on $X$ with respect to $\family {f_i}_{i \mathop \in I}$.

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

Proof
Define:
 * $\forall i \in I: \vartheta_i = \set {U \subseteq X: \map {f_i^{-1} } U \in \tau_i} \subseteq \powerset X$

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

From the Intersection of Topologies is Topology, it suffices to show, for all $i \in I$, that $\vartheta_i$ is a topology on $X$.

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

Let $\AA \subseteq \vartheta_i$.

Then, by Preimage of Union under Mapping: General Result and by the definition of a topology, it follows that:
 * $\ds f_i^{-1} \sqbrk {\bigcup \AA} = \bigcup_{U \mathop \in \AA} f_i^{-1} \sqbrk U \in \tau_i$

That is, $\ds \bigcup \AA \in \vartheta_i$.

Let $U, V \in \vartheta_i$.

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

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

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

That is, $X \in \vartheta_i$.