Homeomorphic Image of Neighborhood Basis is Neighborhood Basis

Theorem
Let Let $T_\alpha = \struct{S_\alpha, \tau_\alpha}$ and $T_\beta = \struct{S_\beta, \tau_\beta}$ be topological spaces.

Let $\phi: T_\alpha \to T_\beta$ be a homeomorphism.

Let $s \in S_\alpha$.

Let $\NN$ be a neighborhood basis of $s$ in $T_\alpha$.

Then:
 * $\NN' = \set{ \phi \sqbrk N : N \in \NN}$ is a neighborhood basis of $\map \phi s$ in $T_\beta$

Proof
Let $N$ be a neighborhood of $s$ in $T_\alpha$.

By definition of neighborhood:
 * $\exists V \in \tau_\alpha : s \in V \subseteq N$

By definition of image of subset:
 * $\map \phi s \in \phi \sqbrk V$

From Subset Maps to Subset:
 * $\phi \sqbrk V \subseteq \phi \sqbrk N$

By definition of homeomorphism:
 * $\phi \sqbrk V \in \tau_\beta$

It follows that $\phi \sqbrk N$ is a neighborhood of $\map \phi s$ in $T_\beta$.

Hence:
 * $\NN' = \set{ \phi \sqbrk N : N \in \NN}$ is a set of neighborhoods of $\map \phi s$ in $T_\beta$

Let $M$ be a neighborhood of $\map \phi s$ in $T_\beta$.

By definition of neighborhood:
 * $\exists U \in \tau_\beta : \map \phi s \in U \subseteq M$

By definition of homeomorphism:
 * $\phi^{-1} \sqbrk U \in \tau_\alpha$ containing $s$

By definition of neighborhood basis:
 * $\exists N \in \NN: s \in N \subseteq \phi^{-1} \sqbrk U$

By definition of image of subset:
 * $\map \phi s \in \phi \sqbrk N$

From Subset Maps to Subset:
 * $\phi \sqbrk N \subseteq \phi \sqbrk {\phi^{-1} \sqbrk U}$

From Image of Preimage under Mapping:
 * $\phi \sqbrk {\phi^{-1} \sqbrk U} \subseteq U$

From Subset Relation is Transitive:
 * $\phi \sqbrk N \subseteq M$

Hence:
 * $\exists \phi \sqbrk N \in \BB': \map \phi s \in \phi \sqbrk N \subseteq M$

The result follows.