Let $(X,d)$ be a metric space. Given a metric space $(Y,\rho)$ and a $1$-Lipschitz map $\varphi\colon X\to Y$, we say that a $1$-Lipschitz function $f\colon X\to \mathbb{R}$ factors through $\varphi$ with a dilation constant $D>0$ whenever there exists a $D$-Lipschitz function $g\colon Y\to \mathbb{R}$ such that $f=g\circ \varphi$.
My question is as follows. When one can guarantee, in terms of properties of the metric space $(X,d)$, the existence of $N\in \mathbb{N}$, $D>0$, and a $1$-Lipschitz map $\varphi \colon X\to \mathbb{R}^N$, where the space $\mathbb{R}^N$ is endowed with, for instance, the norm $\|\cdot \|_{\infty}$, such that the following "universal property" holds: any $1$-Lipschitz function $f\colon X\to \mathbb{R}$ factors through $\varphi$ with the dilation constant $D$.
Clearly, such a property holds in a trivial way as soon as $(X,d)$ is biLipschitz embeddable into a Euclidean space. But nothing comes to my mind besides this. Is this the only case or there is some weaker condition?
Would be grateful for any help.
Okey, unfortunetally, I came to a negative for me answer: any such a universal map is necessarily a biLipschitz embedding. The proof is more or less elementary.
Let me prove a more general statement: suppose for metric spaces $(X,d)$ and $(Y,\rho)$ and a map $\varphi\colon X\to Y$ there holds a property that for any Lipschitz function $f\colon X\to \mathbb{R}$ there is a Lipschitz function $g\colon Y\to \mathbb{R}$ with $g\circ \varphi= f$. Then $\varphi$ is injective and its inverse map $\varphi^{-1}$ defined on its image $\varphi(X)$ is Lipschitz. I removed here the normalization of Lipschitz constants as in fact irrelevant.
First I show that $\varphi$ is injective. Suppose there are $x_1,x_2\in X$ such that $x_1\neq x_2$ but $\varphi(x_1)=\varphi(x_2)$. Clearly, there is a $1$-Lipschitz function $f\colon X\to \mathbb{R}$ with $f(x_1)\neq f(x_2)$. It follows easily from the McShane-Whitney extension lemma. Thus, if for some function $g\colon Y\to \mathbb{R}$ we had $g\circ \varphi =f$, then we would have $f(x_1)=g(\varphi(x_1))=g(\varphi(x_2))=f(x_2)$, which is a contradiction. So, $\varphi$ is indeed injective.
Let $\psi\colon \varphi(X)\to X$ denote the inverse of $\varphi$, so $\psi= \varphi^{-1}$. Now my statement is reduced to the following one: if the post-composition of a map between metric spaces with any real-valued Lipschitz function is Lipschitz, then the map itself is Lipschitz. To prove this, I want to use a statement verified in Lipschitz map between metric and normed spaces. It is shown there that if the post-composition of a map from a metric space to a normed linear space with any continuous linear functional is Lipschitz, then the map itself is Lipschitz.
To reduce my statement to the one obtained in the mentioned post, I use the following standard fact: any metric space can be isometrically embedded into a Banach space. Let $F\colon X \to \mathbb{B}$ be such an embedding with $\mathbb{B}$ being a Banach space. Let $\lambda\in \mathbb{B}^*$ be an arbitrary linear continuous functional on $\mathbb{B}$. The map $\lambda\circ F$ is definitely Lipschitz, so by the assumption I made, the map $\lambda\circ F\circ \psi$ is Lipschitz. Since $\lambda $ is arbitrary, using the fact from the post, I come to the fact that $F\circ \varphi$ is Lipschitz. Since $F$ is an isometric embedding, the map $\varphi$ is also Lipschitz. Q. E. D.