Correspondence between Neighborhood Space and Topological Space

Theorem
Let $S$ be a set.

Let $\struct {S, \tau}$ be a topological space.

Let $\struct {S, \NN}$ be the neighborhood space induced by $\tau$ on $S$.

Let $\phi: \struct {S, \tau} \to \struct {S, \NN}$ be the mapping defined as:
 * $\forall x \in S: \map \phi x = x$
 * $\forall U \in \tau: \phi \sqbrk U = U \in \NN$

Let $\struct {T, \NN'}$ be a neighborhood space.

Let $\struct {T, \tau'}$ be the topological space induced by $\NN$ on $S$.

Let $\psi: \struct {T, \NN'} \to \struct {T, \tau'}$ be the mapping defined as:
 * $\forall y \in T: \map \psi y = y$
 * $\forall V \in \NN': \psi \sqbrk V = V \in \tau'$

Then:
 * $(1): \quad \phi^{-1} \sqbrk {\phi \sqbrk {\struct {S, \tau} } } = \struct {S, \tau}$
 * $(2): \quad \psi^{-1} \sqbrk {\psi \sqbrk {\struct {T, \NN'} } } = \struct {T, \NN'}$

Proof
From the construction of:
 * the neighborhood space induced by $\tau$ on $S$
 * the topological space induced by $\NN$ on $S$

the mappings $\phi$ and $\psi$ are well-defined mappings.

From Topological Space induced by Neighborhood Space induced by Topological Space, $\phi$ is a bijection.

From Neighborhood Space induced by Topological Space induced by Neighborhood Space, $\psi$ is a bijection.

Hence the result.