Equivalence of Definitions of Final Topology/Definition 1 Implies Definition 2
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:
- $\tau = \set{U \subseteq X: \forall i \in I: \map {f_i^{-1}} U \in \tau_i} \subseteq \powerset X$
Then:
- $\tau$ is the finest topology on $X$ such that each $f_i: Y_i \to X$ is $\tuple{\tau_i, \tau}$-continuous.
Proof
From Final Topology is Topology, $\tau$ is a topology.
Mappings are continuous
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.
$\Box$
$\tau$ is the finest such topology
Let $\struct{X, \vartheta}$ be a topological space.
Let the mappings $\family {f_i: Y_i \to X}_{i \mathop \in I}$ be $\tuple {\tau_i, \vartheta}$-continuous.
Let $U \in \vartheta$.
By definition of continuity, for every $i \in I$:
- $\map {f_i^{-1} } U \in \tau_i$
Then by definition of $\tau$:
- $U \in \tau$
It follows that:
- $\vartheta \subseteq \tau$
That is, $\tau$ is finer than $\vartheta$.
$\blacksquare$