Identification Topology is Topology

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Theorem

Let $T_1 = \left({S_1, \tau_1}\right)$ be a topological space.

Let $S_2$ be a set.

Let $f: S_1 \to S_2$ be a mapping.

Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\left({S_1, \tau_1}\right)$.


Then $\tau_2$ is a topology on $S_2$.


Proof

By definition:

$\tau_2 = \left\{{V \in \mathcal P \left({S_2}\right): f^{-1} \left({V}\right) \in \tau_1}\right\}$


We examine each of the open set axioms in turn:


$(O1)$: Union of Open Sets

Let $\left \langle{U_i}\right \rangle_{i \in I}$ be an indexed family of elements of $\tau_2$.

Let $\displaystyle V = \bigcup_{i \mathop \in I} U_i$ be the union of $\left \langle{U_i}\right \rangle_{i \in I}$.

From Preimage of Union under Mapping: Family of Sets:

$\displaystyle f^{-1} \left({\bigcup_{i \mathop \in I} U_i}\right) = \bigcup_{i \mathop \in I} f^{-1} \left({U_i}\right)$

By hypothesis $f^{-1} \left({U_i}\right) \in \tau_1$.

As $\tau_1$ is a topology:

$\displaystyle \bigcup_{i \mathop \in I} f^{-1} \left({U_i}\right) \in \tau_1$

So $f^{-1} \left({V}\right) \in \tau_1$.

Thus $V \in \tau_2$ and so $V$ is open by definition.

$\Box$


$(O2)$: Intersection of Open Sets

Let $U$ and $V$ be elements of $\tau_2$.

From Preimage of Intersection under Mapping:

$f^{-1} \left({U \cap V}\right) = f^{-1} \left({U}\right) \cap f^{-1} \left({V}\right)$

As $\tau_1$ is a topology:

$f^{-1} \left({U}\right) \cap f^{-1} \left({V}\right) \in \tau_1$

Hence $U \cap V$ is open by definition.

$\Box$


$(O3)$: Set Itself

As $f$ is a mapping its domain is $S_1$.

That is:

$f^{-1} \left({S_2}\right) = S_1$

By definition of a topology, $S_1$ is open in $T_1$.

Thus by definition $S_2 \in \tau_2$.

$\Box$


All the open set axioms are fulfilled, and the result follows.

$\blacksquare$


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