Initial Topology with respect to Mapping equals Set of Preimages

Theorem
Let $X$ be a set.

Let $\struct {Y, \tau_Y}$ 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:
 * $\tau_X = \set {f^{-1} \sqbrk U: U \in \tau_Y}$

Proof
Define:
 * $\tau = \set {f^{-1} \sqbrk U: U \in \tau_Y}$

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 $X$.

$({\text O 1})$: Union of Open Sets
Let $\AA \subseteq \tau$.

It is to be shown that:
 * $\ds \bigcup \AA \in \tau$

Define:
 * $\ds \AA' = \set {V \in \tau_Y: f^{-1} \sqbrk V \subseteq \bigcup \AA} \subseteq \tau_Y$

Let:
 * $\ds U = \bigcup \AA'$

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:
 * $\ds f^{-1} \sqbrk U = \bigcup_{V \mathop \in \AA'} f^{-1} \sqbrk V \subseteq \bigcup \AA$

By the definition of $\tau$ and by Set is Subset of Union: General Result, we have:
 * $\ds \forall S \in \AA: \exists V \in \tau_Y: S = f^{-1} \sqbrk V \subseteq \bigcup \AA$

That is:
 * $\forall S \in \AA: \exists V \in \AA': S = f^{-1} \sqbrk V$

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

By, it follows that:
 * $\forall S \in \AA: S \subseteq f^{-1} \sqbrk U$

By Union is Smallest Superset: General Result, we conclude that:
 * $\ds \bigcup \AA \subseteq f^{-1} \sqbrk U$

Hence, by definition of set equality:
 * $\ds \bigcup \AA = f^{-1} \sqbrk U \in \tau$

$({\text O 2})$: Pairwise Intersection of Open Sets
Let $A, B \in \tau$.

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

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

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

$({\text O 3})$: Set Itself
By the definition of a topology, we have $Y \in \tau_Y$.

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