Initial Topology with respect to Mapping equals Set of Preimages

Theorem
Let $X$ be a set.

Let $\left({Y, \tau_Y}\right)$ be a topological space.

Let $f: X \to Y$ be a mapping.

Let $\tau_X$ be the initial topology on $X$ with respect to $f$.

Then:
 * $\tau_X = \left\{{f^{-1} \left({U}\right): U \in \tau_Y}\right\}$

Proof
Define:
 * $\tau = \left\{{f^{-1} \left({U}\right): U \in \tau_Y}\right\}$

By definition, $\tau_X$ is the topology generated by $\tau$.

Therefore, $\tau \subseteq \tau_X$.

If $\tau$ is a topology on $X$, then it follows from the definition of the generated topology that $\tau_X \subseteq \tau$.

By definition of set equality:
 * $\tau_X = \tau$

Hence, it suffices to prove that $\tau$ is a topology on $X$.

We now verify the open set axioms for $\tau$ to be a topology on $X$.

$({O1})$: Union of Open Sets
Let $\mathcal A \subseteq \tau$.

It is to be shown that:
 * $\displaystyle \bigcup \mathcal A \in \tau$

Define:
 * $\displaystyle \mathcal A' = \left\{{V \in \tau_Y: f^{-1} \left({V}\right) \subseteq \bigcup \mathcal A}\right\} \subseteq \tau_Y$

Let:
 * $\displaystyle U = \bigcup \mathcal A'$

By the definition of a topology, we have $U \in \tau_Y$.

By Preimage of Union under Mapping: General Result and Union is Smallest Superset: Family of Sets:
 * $\displaystyle f^{-1} \left({U}\right) = \bigcup_{V \mathop \in \mathcal A'} f^{-1} \left({V}\right) \subseteq \bigcup \mathcal A$

By the definition of $\tau$ and by Set is Subset of Union: General Result, we have:
 * $\displaystyle \forall S \in \mathcal A: \exists V \in \tau_Y: S = f^{-1} \left({V}\right) \subseteq \bigcup \mathcal A$

That is:
 * $\forall S \in \mathcal A: \exists V \in \mathcal A': S = f^{-1} \left({V}\right)$

By Set is Subset of Union: General Result, we have:
 * $\forall V \in \mathcal A': V \subseteq U$

By, it follows that:
 * $\forall S \in \mathcal A: S \subseteq f^{-1} \left({U}\right)$

By Union is Smallest Superset: General Result, we conclude that:
 * $\displaystyle \bigcup \mathcal A \subseteq f^{-1} \left({U}\right)$

Hence, by definition of set equality:
 * $\displaystyle \bigcup \mathcal A = f^{-1} \left({U}\right) \in \tau$

$({O2})$: Pairwise Intersection of Open Sets
Let $A, B \in \tau$.

Let $U, V \in \tau_Y$ be such that $A = f^{-1} \left({U}\right)$ and $B = f^{-1} \left({V}\right)$.

By the definition of a topology, we have $U \cap V \in \tau_Y$.

Then, by Preimage of Intersection under Mapping, $A \cap B = f^{-1} \left({U \cap V}\right) \in \tau$.

$({O3})$: Set Itself
By the definition of a topology, we have $Y \in \tau_Y$.

Hence, by Preimage of Mapping equals Domain, it follows that $X = f^{-1} \left({Y}\right) \in \tau$.