# 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 if and only if 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 Continuity of Composite Mapping, 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:

\(\displaystyle \gamma_1\) | \(\equiv\) | \(\displaystyle \gamma_1 * c\) | here, $\gamma_1 * c$ denotes the concatenation of $\gamma_1$ and $c$ | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle \gamma_1 * \paren {-\gamma_2 * \gamma_2 }\) | by assumption, as $\paren {-\gamma_2 * \gamma_2 }$ and $c$ are both closed paths with initial point $\map {\gamma_1} 0$ | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle \paren {\gamma_1 * \paren {-\gamma_2} } * \gamma_2\) | |||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle c * \gamma_2\) | by assumption | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle \gamma_2\) |

$\Box$

### 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.

$\blacksquare$

## Sources

- 2000: James R. Munkres:
*Topology*(2nd ed.): $\S 52$