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

Let:
 * $\SS = \set{f_i^{-1} \left[{U}\right]: i \in I, U \in \tau_i} \subseteq \map \PP X$

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

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

Then:
 * $\tau$ is the coarsest topology on $X$ such that each $f_i: X \to Y_i$ is $\left({\tau, \tau_i}\right)$-continuous.

Mappings are continuous
Let $i \in I$.

Let $U \in \tau_i$.

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

$\tau$ is the coarsest such topology
Suppose that the mappings are continuous from $\struct{X, \vartheta}$.

Let $U$ be a member of the subbase from Definition 1.

Then for some $i \in I$ and some $V \in \tau_i$,
 * $U = \map {f^{-1}} {V}$

Then since the mappings are continuous from $\struct{X, \vartheta}$:
 * $U \in \vartheta$

Since $\upsilon$ is a topology containing a subbase of $\tau$, $\tau$ is coarser than $\vartheta$.