If $T$ is a nonlinear surjective isometry from Lipschitz-free space $\mathcal{F}(M)$ to $\mathcal{F}(N)$($M,N$ are metric spaces), is $M$ homeomorphic to $N$?



Banach-Stone Theorem in Lipschitz-free spaces
No, this is false --- trivially, because if $M$ is the completion of $N$ then $M$ and $N$ have the same Arens-Eells space. But if you require $M$ and $N$ to both be complete there is a more interesting counterexample.


Let $M$ be three copies of the interval $[0,1]$, joined at the $0$'s, with path metric. (That is, a capital "Y" with path metric.) Let $N$ just be a single copy of the interval $[0,3]$. Then ${\rm Lip}\_0(M)$ and ${\rm Lip}\_0(N)$ are, respectively, isometrically isomorphic to $L^\infty(M)$ and $L^\infty(N)$, and are therefore isometrically isomorphic to each other, by cutting $N$ into three pieces and matching up. (To go from $L^\infty(M)$ to ${\rm Lip}\_0(M)$, integrate from the origin.) The Arens-Eells spaces of $M$ and $N$ sit inside of the dual spaces ${\rm Lip}\_0(M)^\*$ and ${\rm Lip}\_0(N)^\*$ and it is straightforward to check that the dual isometric isomorphism between these spaces restricts to an isometric isomorphism of $AE(M)$ and $AE(N)$.


(If you work through this example you will see how to define the map between $AE(M)$ and $AE(N)$ directly, but I think the explanation I gave above is more illuminating as to why this isometric isomorphism exists.)


On the positive side, there are some good Banach-Stone type theorems if you impose additional conditions on $M$ and $N$; see Sections 3.7 and 3.8 of the second edition of my book on Lipschitz Algebras.




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Edit: The "capital Y" example I described in this answer is not new. A. Godard (Tree metrics and their Lipschitz-free spaces, *Proc. AMS* **138** (2010), 4311-4320) proves that the Arens-Eells space of any separable metric tree is isometrically isomorphic to $L^1[0,1]$. Second comment: If $M$ and $N$ are compact and *concave* in the sense of [this question](https://mathoverflow.net/questions/232337/term-for-a-metric-space-for-which-the-triangle-inequality-is-strict) and $AE(M)$ is isometrically isomorphic to $AE(N)$, then $M$ and $N$ are isometric up to a dilation. (Compactness is not needed if one assumes a uniform version of concavity.) This is Theorem 3.55 of the second edition of my book.


