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

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

Let $\tau$ be the coarsest topology on $X$ such that each $f_i: X \to Y_i$ is $\tuple{\tau, \tau_i}$-continuous.

Let:

$\SS = \set{\map {f_i^{-1}} U: i \in I, U \in \tau_i} \subseteq \map \PP X$

where $\map {f_i^{-1}} U$ denotes the preimage of $U$ under $f_i$.


Then:

$\tau$ is the topology on $X$ generated by the subbase $\SS$.


Proof

Let $\map \tau \SS$ be the topology on $X$ generated by the subbase $\SS$.

$\tau$ contains topology generated by $\SS$

Let $i \in I$ and $U \in \tau_i$.

By definition of $\tuple{\tau, \tau_i}$-continuity:

$\map {f_i^{-1}} U \in \tau$

Since $i \in I$ and $U \in \tau_i$ were arbitrary, then $\SS \subseteq \tau$.

From Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis, $\map \tau \SS \subseteq \tau$

$\Box$

Topology generated by $\SS$ contains $\tau$

Let $i \in I$.

Let $U \in \tau_i$.

Then $\map {f_i^{-1}} {U}$ is an element of the subbase $\SS$ of $X$, and is therefore trivially in $\map \tau \SS$.

By definition of $\tuple{\map \tau \SS, \tau_i}$-continuity:

each $f_i: X \to Y_i$ is $\tuple{\map \tau \SS, \tau_i}$-continuous

Since $\tau$ is the coarsest topology on $X$ such that each $f_i: X \to Y_i$ is $\tuple{\tau, \tau_i}$-continuous then $\tau \subseteq \map \tau \SS$.

By definition of set equality, $\tau = \map \tau \SS$.

$\blacksquare$