Equivalence of Definitions of Initial Topology

Theorem
Let $X$ be a set.

Let $I$ be an indexing set.

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

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

The following definitions of the initial topology on $X$ with respect to $\left \langle {f_i}\right \rangle_{i \mathop \in I}$ are equivalent:

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 $\left({X, \upsilon}\right)$.

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

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

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

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