Is every submetry 1-Lipschitz?

Question

A submetry is usually defined as a function $f:X\to Y$ between metric spaces such that, if $B(x,r)$ is the closed ball of radius $r$, we have the for every $x$ in $X$, $$ f\big(B(x,r)\big) = B \big( f(x),r \big) . $$ Now, some references, such as the link above, require moreover the map $f$ to be 1-Lipschitz (a.k.a. short), while some other references do not. Does that requirement make a difference? That is, is the condition on balls given above sufficient to make $f$ 1-Lipschitz?

Answer 1

If the hypothesis is for all $r>0$, then yes, $f$ is $1$-Lipschitz.

Let $f\colon (X,d_X) \to (Y,d_Y)$ be such that $\forall r> 0$, $f(B_X(x,r)) = B_Y(f(x),r)$. Fix $x\neq z \in X$. Then $d_X(x,z)>0$ and $z \in B(x,d_X(x,z))$ (closed ball). It follows from the hypothesis that $f(z) \in B(f(x),d_X(x,z))$. Therefore, $d_Y(f(x),f(z)) \leqslant d_X(x,z)$, which says that $f$ is 1-Lipschitz.

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Comment: I did not know this kind of objects, but it seems that submetries are defined as 1-Lipschitz functions that are "maximal" in the sense that, OK, they send balls of radius $r$ into balls of radius $r$, as does every 1-Lipschitz function, but here they do it "maximally": they fill these balls. Hence, it is not surprising that some authors mention that they are $1$-Lipschitz. They are meant to form a subset of $1$-Lipschitz functions with a maximality property.