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:
- $\tau \subseteq \tau'$
$\Box$
$\tau$ 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:
- $\tau = \tau'$
$\blacksquare$