Let $Y$ be a complete metric space (I'm interested in particular in the case $Y=c_0$, if relevant) and $X\subset Y$ dense. Given a 1-Lipschitz bijection $f$ from $X$ onto $X$, does it extend to a 1-Lipschitz map $f':Y\rightarrow Y$ which is injective and/or surjective?
Lipschitz maps preserves Cauchy sequences, so the extension exists, but is it possible to recover some injectivity/surjectivity of the map?
I believe the answer might be negative, because for general metric spaces it is negative, but in this case the initial map is bijective from the dense to itself and I'm not sure if this extra information can help somehow.
Best Answer
Here is the example I had in mind in my comment. It is similar to the proposals by Jochen and AnneBauval (from the comments), but is more complicated (and actually works).
For each integer $n$ define the interval $I_n=[n, n+1)$. Set $$ X:= \bigcup_{n\in 2{\mathbb Z}} I_n. $$ The closure $Y$ of $X$ in ${\mathbb R}$ is a disjoint union of closed unit intervals of the form $[2k, 2k+1]$, for integer values of $k$. I will equip $Y$ with the restriction of the standard metric on ${\mathbb R}$.
I define a map $f: Y\to Y$ as the composition of two functions. The first one is $h: x\mapsto x/2$. This function will map $Y$ to the disjoint union of closed intervals of length $1/2$ (and integer left end-point). The other function, $g$, will fix each interval $[n, n+1/2]$ point-wise if $n$ if even and will translate the interval to the left by $1/2$ if $n$ is odd. Thus, $$ g([2k, 2k+1/2)\cup [2k+1, 2k+1.5))=I_{2k}. $$ Moreover, $f=g\circ h$ sends $X$ bijectively to itself. It is easy to see that $f: X\to X$ is 1-Lipschitz. At the same time, $f: Y\to Y$ is not 1-1 (but is surjective).