Relation between homeomorphism and topological equivalence Please consider the following definitions: A: Suppose $d$ and $e$ are metrics on a set $X$. Then, $d$ and $e$ are topologically equivalent metrics if and only if the identity functions from $(X,d)$ to $(X,e)$ and from $(X,e)$ to $(X,d)$ are both continuous.
B: Suppose $(X,d)$ and $(Y,e)$ are metric spaces. Then $X$ and $Y$ are said to be homeomorphic or topologically equivalent if and only if, there exists a bijective function $f:X \rightarrow Y$ that is continuous and has continuous inverse; such a function is called a homeomorphism.

C: Suppose $d$ and $e$ are metrics on a set $X$. If $d$ and $e$ are topologically equivalent, then certainly $(X,d)$ and $(X,e)$ are homeomorphic, because the identity function $I_{d,e}:(X,d) ]\rightarrow (X,e)$ is continuous and has a continuous inverse $I_{e,d}:(X,e) \rightarrow (X,d)$. However, the converse need not be true

In definition C, it is specifically mentioned that the converse need not be true. I assume, by the word : converse, the author implies : if $(X,d)$ and $(X,e)$ are homeomorphic, then,  $d$ and $e$ are topologically equivalent. 
I reason this as follows: there can exist bijective function $f:X \rightarrow Y$ that is continuous and has continuous inverse; but which is not the identity function. Hence, $d$ and $e$ need not be topologically equivalent even though $(X,d)$ and $(X,e)$ are homeomorphic.
Is this correct?
Thanks
A: Yes, this is correct. The "topological equivalence" of metrics only says that they induce the exact same topology.
Here's an example of a scenario the authors thinks about. Let $(X, d)$ and $(X, e)$ be two topologically not-equivalent metrics on a single space $X$ -- for example, let $X = \mathbb{R}^2$, $d$ is standard metric, and $e$ be the star metric: $e(x, y) = d(x, y)$ if $x$ and $y$ lie on the same straight line through origin, and $d(x, 0) + d(0, y)$ if they don't. 
Consider now set $Y = X \coprod X$, that is, the disjoint union of two copies of $X$. Each element of $X$ appears twice in $Y$, we denote these two copies by $x^1$ and $x^2$. We can put two different metrics on $Y$ -- we can have a metric $d_1$ that is equal to $d$ on first copy, and to $e$ on the second copy, and always gives distance $1$ for points in two different copies. We can also have metric $d_2$ that is $e$ on first copy, and $d$ on second copy.
Then, the spaces $(Y, d_1)$ and $(Y, d_2)$ are homeomorphic (actually isometric), homeomorphism just swaps points in copies. However, the identity functions is not homeomorphism.
A: Yes, your statement of the converse is correct.  To construct a counterexample, you might start with a space that admits an involution (order two symmetry), such as the plane and the involution being reflection in the $x$-axis.  Put inequivalent metrics on the two halves such that they agree on the common overlap; call the resulting metric $d$.  Then switch the two; call the resulting metric $e$, i.e. $e(x,y) = d(i(x), i(y))$, where $i:X \to X$ is the involution.  The identity will not be a homeomorphism, but the involution will.  I can provide two such metrics on the two half-planes if you like, or you can think about how to construct one.  Or maybe there is an easier example.
