Equivalence of Definitions of Initial Topology

Theorem
Let $X$ be a set.

Let $I$ be an indexing set.

Let $\left\langle {\left({Y_i, \tau_i}\right)} \right\rangle_{i \in I}$ be an $I$-indexed family of topological spaces.

Let $\left\langle {f_i: X \to Y_i} \right\rangle_{i \in I}$ be an $I$-indexed family of mappings.

Then the initial topology on $X$ with respect to $\left \langle {f_i}\right \rangle_{i \in I}$ by Initial Topology/Definition 1 is the same as the one by Definition:Initial Topology/Definition 2

Proof
As Definition:Initial Topology/Definition 2 implies uniqueness, we need only show that the topology defined by Definition:Initial Topology/Definition 1 satisfies the requirements of Definition:Initial Topology/Definition 2.

Mappings are continuous in definition 1
Let $i \in I$.

Let $U \in \tau_i$.

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

Definition 1 provides the coarsest such topology
Suppose that the mappings are continuous from $(X,\upsilon)$.

Let $U$ be an open set by definition 1.

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

Then since the mappings are continuous from $(X,\upsilon)$:
 * $U \in \upsilon$

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