Skip to main content
added 721 characters in body
Source Link
Piotr Hajlasz
  • 28k
  • 5
  • 86
  • 185

The answer is no. You can have $X$ path connected and $Y$ totally disconnected satisfying your condition. Then any continuous map from $X$ to $Y$ is necessarily constant so there are no injective maps from $X$ to $Y$.

Here is another example: $X=S^1$ is a unit circle of with the geodesic metric and $Y=\mathbb{R}$. Since $X$ and $Y$ are locally isometric, it follows that $$ \kappa_X(\epsilon)\leq \kappa_Y(\epsilon) ; (\forall \epsilon \in (0,1]). $$ Indeed, if $r\le\pi$ in the definition of $\kappa$ the corresponding quantities are equal (balls of radius $r$ are isometric balls). If $r>\pi$, then the quantity in the definition of $\kappa_X(\epsilon)$ is less than the corresponding one in the definition of $\kappa_Y(\epsilon)$ (ball in $Y$ is bigger - segment of length $2r$ than the ball in $S^1$ - the unit circle and they both are locally isometric).

However there is no injective map from $X=S^1$ to $Y=\mathbb{R}$.

The answer is no. You can have $X$ path connected and $Y$ totally disconnected satisfying your condition. Then any continuous map from $X$ to $Y$ is necessarily constant so there are no injective maps from $X$ to $Y$.

The answer is no. You can have $X$ path connected and $Y$ totally disconnected satisfying your condition. Then any continuous map from $X$ to $Y$ is necessarily constant so there are no injective maps from $X$ to $Y$.

Here is another example: $X=S^1$ is a unit circle of with the geodesic metric and $Y=\mathbb{R}$. Since $X$ and $Y$ are locally isometric, it follows that $$ \kappa_X(\epsilon)\leq \kappa_Y(\epsilon) ; (\forall \epsilon \in (0,1]). $$ Indeed, if $r\le\pi$ in the definition of $\kappa$ the corresponding quantities are equal (balls of radius $r$ are isometric balls). If $r>\pi$, then the quantity in the definition of $\kappa_X(\epsilon)$ is less than the corresponding one in the definition of $\kappa_Y(\epsilon)$ (ball in $Y$ is bigger - segment of length $2r$ than the ball in $S^1$ - the unit circle and they both are locally isometric).

However there is no injective map from $X=S^1$ to $Y=\mathbb{R}$.

Source Link
Piotr Hajlasz
  • 28k
  • 5
  • 86
  • 185

The answer is no. You can have $X$ path connected and $Y$ totally disconnected satisfying your condition. Then any continuous map from $X$ to $Y$ is necessarily constant so there are no injective maps from $X$ to $Y$.