Equivalence of Definitions of Final Topology/Definition 2 Implies Definition 1

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 indexed family of topological spaces indexed by $I$.

Let $\family {f_i: Y_i \to X}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$. Let $\tau$ be the finest topology on $X$ such that each $f_i: Y_i \to X$ is $\tuple{\tau_i, \tau}$-continuous.

Then:
 * $\tau = \set{U \subseteq X: \forall i \in I: \map {f_i^{-1}} U \in \tau_i}$

Proof
Let $\tau' = \set{U \subseteq X: \forall i \in I: \map {f_i^{-1}} U \in \tau_i}$.

$\tau'$ contains $\tau$
Let $U \in \tau$.

By definition of $\tuple{\tau_i, \tau}$-continuity for each $i \in I$:
 * $\forall i \in I : \map {f_i^{-1}} U \in \tau_i$

So:
 * $U \in \tau'$.

Since $U$ was arbitrary:
 * $\tau \subseteq \tau'$.

$\tau$ contains $\tau'$
From Final Topology is Topology then $\tau'$ is a topology.

Let $U \in \tau'$.

Let $i \in I$.

Then $\map {f_i^{-1}} {U} \in \tau_i$ by definition of $\tau'$.

It follows that for each $i \in I$, $f_i: Y_i \to X$ is $\tuple{\tau_i, \tau'}$-continuous.

So $\tau'$ is a topology on $X$ such that each $f_i: Y_i \to X$ is $\tuple{\tau_i, \tau'}$-continuous.

Since $\tau$ is the finest topology on $X$ such that each $f_i: Y_i \to X$ is $\tuple{\tau_i, \tau}$-continuous then:
 * $\tau' \subseteq \tau$

By definition of set equality:
 * $\tau = \tau'$.