Suppose $f:C\to C$ is a homeomorphism, where $C=\{0,1\}^{\mathbb N}$ is Cantor space. Suppose $f$ preserves $=^*$ (equality on all but finitely many coordinates). Does it follow that $f$ also reflects $=^*$? That is, if $$x=^* y \implies f(x)=^* f(y)$$ does it follow that: $$x=^* y \iff f(x)=^* f(y)?$$

Using Axiom of Choice we can show that a bijection of $C$ that preserves $=^*$ need not reflect $=^*$, so the continuity assumptions are not superfluous...

Suppose $f:C\to C$ is a homeomorphism, where $C=\{0,1\}^{\mathbb N}$ is Cantor space. Suppose $f$ preserves $=^*$ (equality on all but finitely many coordinates). Does it follow that $f$ also reflects $=^*$? That is, if $$x=^* y \implies f(x)=^* f(y)$$ does it follow that: $$x=^* y \iff f(x)=^* f(y)?$$

Using Axiom of Choice we can show that a bijection of $C$ that preserves $=^*$ need not reflect $=^*$, so the continuity assumptions are not superfluous.

Suppose $f:C\to C$ is a homeomorphism, where $C=\{0,1\}^{\mathbb N}$ is Cantor space. If $x=^* y$ implies $f(x)=^* f(y)$ (equal on all but finitely many coordinates) then is the same true for $f^{-1}$?

I know it fails without the homeorphism assumption if we use Axiom of Choice, for instance...

Suppose $f:C\to C$ is a homeomorphism, where $C=\{0,1\}^{\mathbb N}$ is Cantor space. Suppose $f$ preserves $=^*$ (equality on all but finitely many coordinates). Does it follow that $f$ also reflects $=^*$? That is, if $$x=^* y \implies f(x)=^* f(y)$$ does it follow that: $$x=^* y \iff f(x)=^* f(y)?$$

Using Axiom of Choice we can show that a bijection of $C$ that preserves $=^*$ need not reflect $=^*$, so the continuity assumptions are not superfluous...