Documentation

FormalConjectures.Mathoverflow.«235893»

Mathoverflow 235893 #

Reference: mathoverflow/235893 asked by user Willie Wong

def Mathoverflow235893.IsConnectedMap {X : Type u_1} {Y : Type u_2} [TopologicalSpace X] [TopologicalSpace Y] (f : XY) :

For topological spaces $X$ and $Y$ we say a function $f : X → Y$ is connected is it sends connected sets to connected sets.

Equations
Instances For

    By a standard result, every continuous map is connected

    A set in $\mathbb{R}$ is connected if and only if it is order-connected and non-empty.

    If $f : \mathbb{R} \to \mathbb{R}$ is a connected bijection, then its inverse is also a connected bijection.

    theorem Mathoverflow235893.isConnectedMap_comp {X : Type u_1} {Y : Type u_2} {Z : Type u_3} [TopologicalSpace X] [TopologicalSpace Y] [TopologicalSpace Z] {f : XY} {g : YZ} (hf : IsConnectedMap f) (hg : IsConnectedMap g) :

    The composition of two connected maps is a connected map.

    A homeomorphism is a connected map.

    If $f : \mathbb{R}^1 \to \mathbb{R}^1$ is a connected bijection, then its inverse is also a connected bijection.

    Assume for $n>1$, $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?

    There exists a connected bijection ℝ → ℝ^2 where the inverse is not connected, proven in mathoverflow/260589 by user Gro-Tsen.