# 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 $X$.

### Open Set Axiom $\paren {\text O 1 }$: 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 $U \in \tau_Y$.

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

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 Preimage of Subset is Subset of Preimage, 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:

$\ds \bigcup \AA = f^{-1} \sqbrk U \in \tau$

$\Box$

### Open Set Axiom $\paren {\text O 2 }$: 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, $A \cap B = f^{-1} \sqbrk {U \cap V} \in \tau$.

$\Box$

### Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology

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

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

$\Box$

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

$\blacksquare$