Domain Topology Contains Initial Topology iff Mappings are Continuous

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Theorem

Let $\struct{Y, \tau}$ be a topological space.

Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be a family of topological spaces.

Let $\family {f_i}_{i \mathop \in I}$ be a family of mappings $f_i : Y \to X_i$.

Let $\tau'$ be the initial topology on $Y$ with respect to $\family {f_i}_{i \mathop \in I}$.


Then:

$\tau' \subseteq \tau$ if and only if $\forall i \in I : f_i: \struct{Y, \tau} \to \struct{X_i, \tau_i}$ is $\tuple{\tau, \tau_i}$-continuous.


Proof

Necessary Condition

Let $\tau' \subseteq \tau$.

From Equivalence of Definitions of Initial Topology:

for each $i \in I$, $f_i: \struct{Y, \tau'} \to \struct{X_i, \tau_i}$ is $\tuple{\tau', \tau_i}$-continuous

From Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous:

for each $i \in I$, $f_i: \struct{Y, \tau} \to \struct{X_i, \tau_i}$ is $\tuple{\tau, \tau_i}$-continuous

$\Box$


Sufficient Condition

For all $i \in I$, let $f_i: \struct{Y, \tau} \to \struct{X_i, \tau_i}$ be $\tuple{\tau, \tau_i}$-continuous.

From Equivalence of Definitions of Initial Topology:

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

Then:

$\tau' \subseteq \tau$

$\blacksquare$