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What does "identity map $id$" mean in this context? Two metrics $d_1$ and $d_2$ on $X$ are said to be Lipschitz equivalent if the identity map $id\colon (X,d_1)\to (X,d_2)$ is bilipschitz.

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up vote 7 down vote accepted

The identity map always means $Id(x)=x$, in every context.

In this one, note that the underlying set $X$ is the same, so there is sense to speak about the identity map. However the metrics may be different, which would imply a different topology.

The identity map need not be continuous when going from one topology to another (e.g. $x_0$ is an isolated point in the range, but not in the domain), but in a sense it measures some compatibility between the topologies and in this case the metrics.

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At least in finite dimensional vector spaces, the norms are equivalent, yielding equivalent topologies. Thus, $Id$ is continuous. – robjohn Nov 9 '11 at 8:29

The symbol 'id' is often used to refer to the identity map, which is the map which takes every element of a set $X$ to itself:

$$\begin{align} \textrm{id} : & X \to X \\ & x \mapsto x \end{align}$$

or, phrased differently, id is the map from $X$ to $X$ that satisfies $\textrm{id}(x)=x$ for every $x\in X$.

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It is the map $X\to X$ given by the formula $x\mapsto x$.

It is the identity map of the underlying set $X$ of the two metric spaces you consider. There is not really an identity map from $(X,d_1)$ to $(X,d_2)$ because these are not the same metric spaces.

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