# Equivalence of Definitions of Initial Topology

## 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 $I$.

The following definitions of the concept of Initial Topology are equivalent:

### Definition 1

Let:

$\mathcal S = \left\{{f_i^{-1} \left[{U}\right]: i \in I, U \in \tau_i}\right\} \subseteq \mathcal P \left({X}\right)$

where $f_i^{-1} \left[{U}\right]$ denotes the preimage of $U$ under $f_i$.

The topology $\tau$ on $X$ generated by $\mathcal S$ is called the initial topology on $X$ with respect to $\left \langle {f_i}\right \rangle_{i \mathop \in I}$.

### Definition 2

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

Then $\tau$ is known as the initial topology on $X$ with respect to $\left \langle {f_i} \right \rangle_{i \mathop \in I}$.

## Proof

### Definition 1 Implies Definition 2

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 $\SS$.

#### Mappings are 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 $\tau$.

$\Box$

#### $\tau$ is the coarsest such topology

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 $f_i$:

$U \in \vartheta$

From Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis, $\tau \in \vartheta$

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

$\Box$

### Definition 2 Implies Definition 1

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 $f_i$.

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

#### $\tau$ contains topology generated by $\SS$

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

By definition of $\tuple{\tau, \tau_i}$-continuity:

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

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

From Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis, $\map \tau \SS \subseteq \tau$

$\Box$

#### Topology generated by $\SS$ contains $\tau$

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 $\map \tau \SS$.

By definition of $\tuple{\map \tau \SS, \tau_i}$-continuity:

each $f_i: X \to Y_i$ is $\tuple{\map \tau \SS, \tau_i}$-continuous

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

By definition of set equality, $\tau = \map \tau \SS$.

$\blacksquare$