Question regarding homeomorphisms If $(X,d)$ and $(Y,D)$ are two metric spaces and let $f$ be a homeomorphism between the two metric spaces. Define a new metric $p$ on $X$ defined by $p(x,y) = D(f(x),f(y))$.Then is it true that $d$ and $p$ are equivalent metrics on $X$? 
 A: They need not be Lipschitz-equivalent, as can be seen by taking $(Y,D) = (X, d/(1+d))$ and $f = {\rm Id}_X$ (as long as $d$ is not bounded). But it appears that $d$ and $p$ will always generate the same topology. 
Note that for all $x \in X$ and $r>0$ we have $B_p(x,r) = f^{-1}(B_D(f(x),r))$.
Fix $x \in X$ and $r> 0$. We want $r' > 0$ such that $B_p(x,r')\subseteq B_d(x,r)$, right? Since homeomorphisms are open mappings, $f(B_d(x,r))$ is an $D$-open set around $f(x)$, so there is $r' > 0$ such that $B_D(f(x),r')\subseteq f(B_d(x,r))$. This implies that $B_p(x,r') \subseteq B_d(x,r)$, as wanted.
For the other direction, fix $x \in X$ and $r>0$. This time we want $r'>0$ such that $B_d(x,r')\subseteq B_p(x,r)$. Note that $B_p(x,r) = f^{-1}(B_D(f(x),r))$ is an $d$-open set around $x$, by continuity of $f$, so we get $r'>0$ such that $B_d(x,r')\subseteq B_p(x,r)$, done.
A: No. Take $X=Y=\mathbb R$, and $f(x)=x^{1/3}$. Then the identity map $\mathrm{Id}: (\mathbb R, D) \to (\mathbb R, p)$ is not bilipschitz, so the two metrics are not equivalent (though they give the same topology).
