Homeomorphic Topology of Initial Topology is Initial Topology

Theorem
Let $\struct{X_\alpha, \tau_\alpha}, \struct{X_\beta, \tau_\beta}$ be topological spaces.

Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.

Let $\family {f_i: X_\beta \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$.

Let $\tau_\beta$ be the initial topology on $X_\beta$ with respect to $\family {f_i}_{i \mathop \in I}$.

Let $\phi : \struct{X_\alpha, \tau_\alpha} \to \struct{X_\beta, \tau_\beta}$ be a homeomorphism.

Then:
 * $\tau_\alpha$ is the initial topology on $X_\alpha$ with respect to $\family {f_i \circ \phi : X_\alpha \to Y_i}_{i \mathop \in I}$

Proof
Let $\SS_\beta = \set {f_i^{-1} \sqbrk U: i \in I, U \in \tau_i}$.

By definition of initial topology:
 * $\SS_\beta$ is a sub-basis for $\tau_\beta$

From Inverse of Homeomorphism is Homeomorphism:
 * $\phi^{-1}$ is a homeomorphism

From User:Leigh.Samphier/Topology/Homeomorphic Image of Sub-Basis is Sub-Basis:
 * $\SS_\alpha = \set{ \phi^{-1} \sqbrk {f_i^{-1} \sqbrk U} : i \in I, U \in \tau_i}$ is a sub-basis for $\tau_\alpha$

From Preimage of Subset under Composite Mapping:
 * $\forall i \in I, U \in \tau_i : \phi^{-1} \sqbrk {f_i^{-1} \sqbrk U} = \paren{f_i \circ \phi}^{-1} \sqbrk U$

Hence:
 * $\SS_\alpha = \set{ \paren{f_i \circ \phi}^{-1} \sqbrk U : i \in I, U \in \tau_i}$ is a sub-basis for $\tau_\alpha$

By definition, $\tau_\alpha$ is the initial topology on $X_\alpha$ with respect to $\family {f_i \circ \phi}_{i \mathop \in I}$