Metrics on Space are Topologically Equivalent iff Identity Mapping is Homemorphism

Theorem
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

Let $I_A$ denote the identity mapping on $A$.

Then:
 * $d_1$ and $d_2$ are topologically equivalent


 * $I_A: M_1 \to M_2$ is a homeomorphism.
 * $I_A: M_1 \to M_2$ is a homeomorphism.

Proof
First we establish the following:

From Identity Mapping is Bijection, we have that $I_A$ is a bijection.

From Inverse of Identity Mapping, $\paren {I_A}^{-1} = I_A$.

By Identity Mapping on Metric Space is Continuous:
 * $(1): \quad I_A$ is a $\tuple {d_1, d_1}$-continuous mapping.
 * $(2): \quad I_A^{-1} = I_A$ is a $\tuple {d_2, d_2}$-continuous mapping.

Sufficient Condition
Let $d_1$ and $d_2$ be topologically equivalent.

Hence, by definition:


 * For all metric spaces $\struct {B, d}$ and $\struct {C, d'}$:
 * For all mappings $f: B \to A$ and $g: A \to C$:


 * $(1): \quad f$ is $\tuple {d, d_1}$-continuous $f$ is $\tuple {d, d_2}$-continuous
 * $(2): \quad g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous.

In particular, for all mappings $f: A \to A$ and $g: A \to A$:


 * $(1): \quad f$ is $\tuple {d_2, d_1}$-continuous $f$ is $\tuple {d_2, d_2}$-continuous
 * $(2): \quad g$ is $\tuple {d_1, d_1}$-continuous $g$ is $\tuple {d_2, d_1}$-continuous.

By definition of topological equivalence:
 * $I_A$ is a $\tuple {d_1, d_2}$-continuous mapping from $(1)$
 * $I_A$ is a $\tuple {d_2, d_1}$-continuous mapping from $(2)$.

Hence, $I_A$ is a bijection such that:
 * $I_A$ is continuous from $M_1$ to $M_2$
 * $I_A^{-1}$ is continuous from $M_2$ to $M_1$.

That is, $I_A$ is a homeomorphism.

Necessary Condition
Let $I_A$ be a homeomorphism.

That is:
 * $I_A$ is continuous from $M_1$ to $M_2$
 * $I_A^{-1}$ is continuous from $M_2$ to $M_1$.

it having been established that $I_A$ is a bijection such that $I_A^{-1} = I_A$.

Let $U \subseteq A$ be open in $M_2$.

Because $I_A$ is a homeomorphism:
 * $I_A^{-1} \sqbrk U$ is open in $M_1$.

Let $U \subseteq A$ be open in $M_1$.

Because $I_A^{-1} = I_A$ is a homeomorphism:
 * $I_A \sqbrk U$ is open in $M_2$.

That is:
 * $U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_2$-open.

That is:
 * $d_1$ and $d_2$ are topologically equivalent.