User:Anghel/Sandbox

Definition
Let $T = \struct{S, \tau}$ be a path-connected topological space.

$T$ is said to be simply connected if the fundamental group $\map {\pi_1}{ T }$ of $T$ is trivial.

$T$ is said to be simply connected if all loops in $T$ with identical base points are path-homotopic.

$T$ is said to be simply connected if all paths in $T$ with identical initial points and final points are path-homotopic.

$T$ is said to be simply connected if all loops in $T$ are path-homotopic with a constant mapping.

Also see

 * Simple Connectedness is Preserved under Homeomorphism, which shows that simple connectedness is a topological property.
 * Fundamental Group is Independent of Base Point for Path-Connected Space

Links

 * Fundamental Group is Independent of Base Point for Path-Connected Space

Category:Definitions/Simply Connected Spaces]] Category:Definitions/Topology]] Category:Definitions/Algebraic Topology]]

Theorem
Let $T = \struct{S, \tau}$ be a path-connected topological space.

1 => 2
Let $x \in S$.

From Fundamental Group is Independent of Base Point for Path-Connected Space, it follows that all fundamental groups $\map {\pi_1}{T, x}$ are isomorphic to one group denoted $\map {\pi_1}{ T }$.

By assumption, it follows that $\map {\pi_1}{ T }$ is trivial.

As the single element of $\map {\pi_1}{T, x}$ is a homotopy class of paths, it follows that all loops in $T$ with base point $x$ belong to the same homotopy class.

It follows that these loops are path-homotopic.

2 => 1
As all loops in $T$ with base point $x \in S$ are path-homotopic, it follows that there exists only one homotopy class of loops with base point $x$.

It follows that the fundamental group $\map {\pi_1}{T, x}$ only has one element, so it is trivial.

From Fundamental Group is Independent of Base Point for Path-Connected Space, it follows that all fundamental groups $\map {\pi_1}{T, x}$ are isomorphic to $\map {\pi_1}{ T }$.

2 => 4
Let $x \in S$, and let $c : \closedint 0 1 \to S$ be the constant mapping defined by $\map c t = x$.

From Constant Mapping is Continuous, it follows that $c$ is continuous.

By definition of loop, it follows that $c$ is a loop with base point $x$.

By assumption, all loops with base point $x$ are path-homotopic with $c$.

Note that this means that all loops in $T$ are nulhomotopic.

4 => 2
Let $\gamma_1, \gamma_2$ be a loops in $T$ with base point $x$.

By assumption, there exists a path-homotopy between $\gamma_1$ and a constant mapping $c_1 : \closedint 0 1 \to S$.

By definition of path-homotopy, it follows that $\map {c_1} 0 = \map {\gamma_1} 0 = x$.

It follows that the constant mapping $c_2 : \closedint 0 1 \to S$ path-homotopic to $\gamma_2$ also have $\map {c_2} 0 = x$, so $c_1$ and $c_2$ are identical.

As Relative Homotopy is Equivalence Relation, and $\gamma_1$ and $\gamma_2$ are both path-homotopic to $c_1$, it follows that $\gamma_1$ and $\gamma_2$ are path-homotopic.

4 => 3
Let $\gamma_1, \gamma_2: \closedint 0 1 \to S$ be two paths with $\map {\gamma_1} 0 = \map {\gamma_2} 0$ and $\map {\gamma_1} 1 = \map {\gamma_2} 1$.

Define the mapping $-\gamma_2: \closedint 0 1 \to X$ by:


 * $-\map {\gamma_2} t = \map {\gamma_2} {1 - t}$.

From Composite of Continuous Mappings is Continuous, it follows that $-\gamma_2$ is a path.

Let $c: \closedint 0 1 \to X$ be the constant mapping defined by $\map c t = \map {\gamma_1} 0$.

When $\equiv$ denotes equivalence of homotopy classes, we have:

3 => 2
Let $\gamma_1, \gamma_2$ be loops in $T$ with base point $x$.

By assumption, $\gamma_1$ and $\gamma_2$ are path-homotopic.

Category:Simply Connected Spaces]]