Equiconnected implies contractible

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., equiconnected space) must also satisfy the second topological space property (i.e., contractible space)
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Statement

Any equiconnected space is contractible.

Definitions used

Equiconnected space

Further information: equiconnected space

A nonempty topological space $X$ is said to be equiconnected if there is a continuous map $k : X \times [0, 1] \times X \to X$ such that $k (x, t, x) = x$ for all $x$ and $k (x, 0, y) = x, k (x, 1, y) = y$ for all $x$ and $y$ .

Contractible space

Further information: contractible space

A nonempty topological space $X$ is said to be contractible if there exists a point $x_{0} \in X$ and a continuous map $F : X \times [0, 1] \to X$ such that $F (x, 0) = x$ for all $x$ , and $F (x, 1) = x_{0}$ for all $x$ . (Another equivalent definition states that this is true for every $x_{0} \in X$ ).

Facts used

Slice inclusion is continuous: For a product of topological space $A \times B$ and a point $b \in B$ , the map $A \to A \times B$ given by $a \mapsto (a, b)$ is continuous.
The composition of continuous maps is continuous

Proof

Given: A nonempty topological space $X$ , a continuous map $k : X \times [0, 1] \times X \to X$ such that $k (x, t, x) = x$ for all $x$ and $k (x, 0, y) = x, k (x, 1, y) = y$ for all $x$ and $y$

To prove: There exists a point $x_{0} \in X$ and a continuous map $F : X \times [0, 1] \to X$ such that $F (x, 0) = x$ for all $x$ , and $F (x, 1) = x_{0}$ for all $x$ .

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Choose any point $x_{0} \in X$		$X$ is nonempty
2	Define $F : X \times [0, 1] \to X$ as $F (x, t) : = k (x, t, x_{0})$ .		$k : X \times [0, 1] \times X \to X$ exists	Step (1)
3	$F$ is continuous.	Facts (1), (2)	$k$ is continuous	Step (2)	$F$ is the composite of $k$ applied on the slice inclusion map $X \times [0, 1] \to X \times [0, 1] \times X$ defined by (x,t) \mapsto (x,t,x_0)</math>. By Fact (1), the slice inclusion is continuous, so by Fact (2), the composite is continuous.
4	$F (x, 0) = x$ for all $x \in X$ .		$k (x, 0, y) = x$ for all $x, y \in X$	Step (2)	$F (x, 0) = k (x, 0, x_{0})$ by Step (2). Combining this with the given data, we get that $F (x, 0) = x$ .
5	$F (x, 1) = x_{0}$ for all $x \in X$ .		$k (x, 1, y) = y$ for all $x, y \in X$	Step (2)	$F (x, 1) = k (x, 1, x_{0})$ by Step (2). Combining this with the given data, we get that $F (x, 1) = x_{0}$ .
6	$x_{0}$ from Step (1) and $F$ from Step (2) are as desired.			Steps (1), (2), (3), (4), (5)	$x_{0} \in X$ by Step (1) and $F$ has the correct domain and co-domain by Step(2). Steps (3), (4), (5) establish the three additional things that need to be established: continuity, value at 0, and value at 1. This completes the proof.

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