Equivalence of Definitions of Metrizable Topology/Lemma 1

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

Let $M = \struct {A, d}$ be a metric space.

Let $\tau_d$ be the topology induced by $d$ on $A$.

Let $\phi : \struct{S, \tau} \to \struct{A, \tau_d}$ be a homeomorphism between topological spaces $\struct{S, \tau}$ and $\struct{A, \tau_d}$.

Let $d_\phi : S \times S \to \R_{\ge 0}$ be the mapping defined by:
 * $\forall s,t \in S: \map {d_\phi} {s,t} = \map d {\map \phi s, \map \phi t}$

Then:
 * $d_\phi$ is a metric on $S$

$d_\phi$ satisfies Metric Axiom $(\text M 1)$
We have:

Hence $d_\phi$ satisfies metric axiom $(\text M 1)$.

$d_\phi$ satisfies Metric Axiom $(\text M 2)$
We have:

Hence $d_\phi$ satisfies metric axiom $(\text M 2)$.

$d_\phi$ satisfies Metric Axiom $(\text M 3)$
We have:

Hence $d_\phi$ satisfies metric axiom $(\text M 3)$.

$d_\phi$ satisfies Metric Axiom $(\text M 4)$
By definition of homeomorphism:
 * $\forall s, t \in S : s \ne t \iff \map \phi s \ne \map \phi t$

We have:

Hence $d_\phi$ satisfies metric axiom $(\text M 4)$.

It follows that $d_\phi$ is a metric on $S$ by definition.