Can one cancel $\mathbb R$ in a bi-Lipschitz embedding?

Question

Let $X$ be a metric space. Suppose that the product $X\times\mathbb R$ admits a bi-Lipschitz embedding into $\mathbb R^{n}$. Does it follow that $X$ admits a bi-Lipschitz embedding into $\mathbb R^{n-1}$?

Definitions and comments:

1. A map $F\colon (Y,\rho)\to\mathbb R^n$ is a bi-Lipschitz embedding if there exists a constant $L$ such that $L^{-1}\rho(a,b)\le |F(a)-F(b)|\le L\rho(a,b)$ for all $a,b\in Y$.
2. The choice of a product metric on $X\times \mathbb R$ makes no difference. For definiteness, let the distance between $(x,t)$ and $(x',t')$ be $d_X(x,x')+|t-t'|$.
3. The cancellation fails if bi-Lipschitz is replaced by topological or smooth embedding. For example, $S^1\times \mathbb R$ embeds into $\mathbb R^2$ smoothly, but $S^1$ does not embed into $\mathbb R$ in any sense.
4. The cancellation is possible when $n=1$, according to $\mathbb R^0=\{0\}$. :) I'm pretty sure it's also possible when $n=2$, but don't have a proof. For $n\ge 3$ I don't have a clue.

Answer 1

I think the answer to your question does not depend at all on the target space, but only on the domain of your function and how you define the metric on it. If you use the metric you suggested, then the answer is positive for whichever target space $Y$. For, suppose $F\colon X\times\mathbb{R}\to Y$ is a bi-Lipschitz embedding into any metric space $Y$. Define a new map $\hat F\colon X\to Y$ by setting $\hat F(x)=F(x,0)$. Then you get: $d_{Y}(\hat F(x),\hat F(y))=d_Y(F(x,0),F(y,0))\le L d_{X\times\mathbb{R}}((x,0),(y,0))=d_X(x,y)$ and similarly for the other inequality. I hope this helps!