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:

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

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

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

Then:

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

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 $\tau$.

$\Box$

Let $\struct {X, \vartheta}$ be a topological space.

Let the mappings $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be $\tuple {\vartheta, \tau_i}$-continuous.

Let $U \in \SS$.

Then for some $i \in I$ and some $V \in \tau_i$:

$U = \map {f_i^{-1} } V$

By definition of the continuity of $f_i$:

$U \in \vartheta$

$\tau \in \vartheta$

That is, $\tau$ is coarser than $\vartheta$.

$\blacksquare$