trouble with understanding the problem statement about metric inducing the topology My problem statement says:
Let $(X,d)$ be a metric space. Assume $f:X \to Y$ is a homeomorphism and define $d^*:Y\times Y \to \mathbb{R}$ by $d^*(x,y)= d(f^{-1}(x), f^{-1}(y))$.
(a) Prove that $d^*$ is a metric on $Y$.
(b) Prove that the topology on $Y$ is induced by the metric $d^*$.
(c) Prove that $f$ is an isometry between $(X,d)$ and $(Y,d^*)$.
I have trouble with part (b), mostly with understanding what exactly the problem wants me to prove. The fact that it says "$\textit{the}$ topology on $Y$ is induced by the metric $d^*$", sounds to me that it says any topology on $Y$ is induced by $d^*$, which is not correct. So I am thinking it probably wants me to prove that the topology induced by $d^*$ is indeed a topology. Am I right? Also, do I need to use $d^*(x,y)=d(x,y) = d(f^{-1}(x), f^{-1}(y))$ for part (b), or is it asking about a generic metric? Any clarifications of what part (b) wants me to prove will be very much appreciated.
 A: $(Y, \mathcal{T})$ is a topological space such that $X$ (in its metric topology from $d$, so $\mathcal{T}_d=\mathcal{T}_X$) is homeomorphic to $Y$ via $f$.
The $d^\ast(x,y) = d(f^{-1}(x), f^{-1}(y))$ (well-defined, because $f$ is a bijection) is indeed a metric on $Y$ (this is what question a) is; check the axioms for a metric) and as such induces a topology on $Y$ as well, $\mathcal{T}_{d^\ast}$. The question under b) then asks to show that $\mathcal{T}_Y=\mathcal{T}_{d^\ast}$. 
For that it will help to note that 
$$\forall x \in X, \forall y \in Y, \forall r>0: f[B_d(x,r)]=B_{d^\ast}(f(x), r) \text{ and } f^{-1}[B_{d^\ast}(y,r)]=B_d(f^{-1}(y), r)$$
to see that $d^\ast$-open balls are open in $\mathcal{T}_Y$ among other things..
Finally c) is just checking the definition of isometry. 
a) and b) together show that "being a metrisable topology" is a topological property: a space homeomorphic to a topology induced by a metric is itself inducable by a metric too.
A: a) (i) $d^*(x, y) = d(f^{-1}(x), f^{-1}(y)) > 0$ for all $x, y \in Y$ since $d$ is a metric, and $d^*(x, y)=0 \iff d(f^{-1}(x), f^{-1}(y))=0 \iff f^{-1}(x)= f^{-1}(y) \iff x=y$, since $f$ is a bijection.
(ii) $d^*(x, y) = d(f^{-1}(x), f^{-1}(y)) = d(f^{-1}(y), f^{-1}(x)) = d^*(y, x)$ for all $x, y \in Y$.
(iii) $d^*(x, y) = d(f^{-1}(x), f^{-1}(y)) \le d(f^{-1}(x), f^{-1}(z)) + d(f^{-1}(z), f^{-1}(y)) = d^*(x, z) + d^*(z, y)$ for all $x, y, z \in Y$.
b) As has been made clear by the comments/answer previously given, since we know that $Y$ is a topological space because $f$ is a homeomorphism, the question is simply asking that we show that the topology induced by the metric $d^*$ is the same as the one topology on $Y$ that we already know of. To do this, we compare the open sets/balls and show they are one and the same.
Let $\epsilon >0$ be arbitrary, $x\in X$ and $y\in Y$. Let $U$ be open in the topological space $Y$ that is homeomorphic to $X$ via $f$. This means that $f^{-1}(U)$ is open in $X$. Thus, $f^{-1}(U)=\bigcup B_d(x, \epsilon)$. Then $U = \bigcup f(B_d(x, \epsilon)) = \bigcup B_{d^*}(f(x), \epsilon)$. This means that the open sets $U$ are made up of the open sets of the metric topology on $Y$, $B_{d^*}(y, \epsilon)$. Now if we consider $B_{d^*}(y, \epsilon)$, open in $Y$ with the metric topology induced by $d^*$, then $f^{-1}(B_{d^*}(y, \epsilon)) = B_{d}(f^{-1}(y), \epsilon)$ implies that $B_{d^*}(y, \epsilon)$ has to be open in the topology on $Y$ coming form $X$ via $f$. We have shown that both of the topologies are the equal.
c) To prove that for all $x_1, x_2 \in X$, $d^*(f(x_1), f(x_2)) = d(x_1, x_2)$, note that $d^*(f(x_1), f(x_2)) = d(f^{-1}(f(x_1)), f^{-1}(f(x_2))) = d(x_1, x_2)$.
