Let $(X,d)$ and $(Y,d')$ be metric spaces and $f : X \to Y$ a homeomorphism. Show that $\rho(x,y)= d'(f(x),f(y))$ defines a metric $\rho$ in $X$ and that $d$ and $\rho$ are equivalent.
To show that $\rho(x,y)$ is a metric I have $$\rho(x,y)=d'(f(x),f(y)) \leqslant d'(f(x),f(y)) + d'(f(y),f(z)) = \rho(x,y)+\rho(y,z) \\ \rho(x,y)=d'(f(x),f(y))=d'(f(y),f(x)) = \rho(y,x) \\ \rho(x,y)=0 \iff d'(f(x),f(y))=0 \implies f(x)=f(y) \underset{f \text{ homeo}}\iff x=y$$
thus $\rho$ is a metric in $X$. To show that $\rho$ and $d$ are equivalent I'm a bit lost. I have three metrics here and I'm not sure where I should even use $d$ since $\rho$ depends on $d'$ not $d$.
In essence I would need to show that $\operatorname{id} :(X,d) \to (Y,d') \to (X,\rho)$ is homeomorphic? The fact that $f$ is homeomorphic implies that there exists and continuous inverse $f^{-1}:(Y,d') \to (X,d)$, but this also says nothing about $(X,\rho)$.
$\textbf{If}$ I would have that $f^{-1}:(Y,d') \to (X,\rho)$, then picking $y_1,y_2 \in Y$ I would have that $$\rho(f^{-1}(y_1), f^{-1}(y_2))=d'(f(f^{-1}(y_1)),f(f^{-1}(y_2)) = d'(y_1,y_2)$$ which would imply that $f^{-1}$ is a bijective isometry from $(Y,d') \to(X,\rho)$ and thus homeomorphism. Furhtermore I would have that $f^{-1} \circ f :(X,d) \to(X,\rho)$ would be a homeo since it's a composition of two homeos. So $\operatorname{id} :(X,d) \to (X,\rho)$ is homeomorphic and $d$ and $\rho$ are equivalent.
I'm very confused about the fact that I have three metrics. Any clarification would be welcome.
Proof 1
For any $x \in X$, $D$ a distance on $X$ and $r \gt 0$, denote by $B_r(x;D)$ the open ball for the distance $D$ centered on $x$ and of radius $r$.
By definition of equivalent distances, we have to prove that for any $x \in X$ and $r \gt 0$, we can find $r_1,r_2 \gt 0$ such that
$$B_{r_1}(x,d) \subseteq B_{r}(x,\rho) \text{ and } B_{r_2}(x,\rho) \subseteq B_{r}(x,d).$$
By definition of the distance $\rho$, we have
$$B_{r}(x,\rho)= \{y \in X \mid d^\prime(f(x),f(y)) \lt r\} .$$ As $f$ is supposed to be an homeomorphism between $(X,d)$ and $(Y,d^\prime)$, it is continuous at $x$ and it exists $r_1 \gt 0$ such that for all $y \in X$ with $d(x,y) \lt r_1$, we have $d^\prime(f(x),f(y)) \lt r$. This exactly means that $B_{r_1}(x,d) \subseteq B_{r}(x,\rho)$.
I let you deal with the case of finding $r_2$ using the continuity of $f^{-1}$.
Proof 2
Consider
$$\operatorname{id} :(X,d) \mathop{\to}^{f} (Y,d^\prime) \mathop{\to}^{f^{-1}} (X,\rho).$$
$\operatorname{id}$ is well defined as $f$ is supposed to be a homeomorphism between $(X,d)$ and $(Y,d^\prime)$.
$f^{-1}$ is an isometric map between $(Y,d^\prime)$ and $(X, \rho)$ as for $x^\prime = f(x), y^\prime =f(y) \in Y$ we have $$\rho(f^{-1}(x^\prime),f^{-1}(y^\prime))=\rho(x,y)=d^\prime(f(x),f(y))= d^\prime(x^\prime,y^\prime)$$
$\operatorname{id}$ which is the composition of two continuous maps (the homeomorphism $f$ and the isometry $f^{-1}$) is $(d,\rho)$-continuous.
In a similar way, we could prove that
$$\operatorname{id} :(X,\rho) \mathop{\to}^{f} (Y,d^\prime) \mathop{\to}^{f^{-1}} (X,d)$$ is $(\rho,d)$-continuous.
Finally $d, \rho$ are equivalent as $\operatorname{id}$ is both $(d, \rho)$ and $(\rho,d)$ continuous.