Homotopy Characterisation of Simply Connected Sets

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

Let $X$ be a subset of $S$.

Then $X$ is simply connected the following conditions hold:


 * $(1): \quad $ $X$ is path-connected.
 * $(2): \quad $ All paths in $X$ with the same initial points and final points are freely homotopic.

Proof
We only have to check condition $(2)$, as a simply connected set is path-connected by definition.

Necessary condition
Suppose that $X$ is simply connected.

Let $\gamma_1, \gamma_2: \closedint 0 1 \to X$ 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 path defined by $\map c t = \map {\gamma_1} 0$.

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

Sufficient Condition
Let $x_0 \in X$.

We show that the fundamental group $\map {\pi_1} {X, x_0}$ is trivial.

Let $\gamma_1, \gamma_2: \closedint 0 1 \to X$ be two closed paths with initial point $x_0$.

By assumption, $\gamma_1$ and $\gamma_2$ are freely homotopic and belong to the same homotopy class.

Hence, $\map {\pi_1} {X, x_0}$ has only one element.