Identification Topology is Topology

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ 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 $\struct {S_1, \tau_1}$.

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

Proof
By definition:
 * $\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$

We examine each of the open set axioms in turn:

Let $\family {U_i}_{i \mathop \in I}$ be an indexed family of elements of $\tau_2$.

Let $\ds V = \bigcup_{i \mathop \in I} U_i$ be the union of $\family {U_i}_{i \mathop \in I}$.

From Preimage of Union under Mapping: Family of Sets:


 * $\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} U_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {U_i}$

We have $f^{-1} \sqbrk {U_i} \in \tau_1$.

As $\tau_1$ is a topology:
 * $\ds \bigcup_{i \mathop \in I} f^{-1} \sqbrk {U_i} \in \tau_1$

So $f^{-1} \sqbrk V \in \tau_1$.

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

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

From Preimage of Intersection under Mapping:


 * $f^{-1} \sqbrk {U \cap V} = f^{-1} \sqbrk U \cap f^{-1} \sqbrk V$

As $\tau_1$ is a topology:
 * $f^{-1} \sqbrk U \cap f^{-1} \sqbrk V \in \tau_1$

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

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

That is:
 * $f^{-1} \sqbrk {S_2} = S_1$

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

Thus by definition $S_2 \in \tau_2$.

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