Equivalence of Definitions of Initial 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 $\family {f_i: X \to Y_i}_{i \mathop \in I}$ 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:

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

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

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

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

$\map \tau \SS \subseteq \tau$

$\Box$

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 $\map \tau \SS$.

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

$\tau \subseteq \map \tau \SS$

By definition of set equality:

$\tau = \map \tau \SS$

$\blacksquare$