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 $U \in \tau_Y$.
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 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.
$\blacksquare$