Star Convex Set is Simply Connected

Theorem
Let $A$ be a star convex subset of a topological vector space $\struct {V, \tau}$ over $\R$ or $\C$.

Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$.

Then $\struct{ A, \tau_A }$ is simply connected.

Proof
Let $a \in A$ be a star center of $A$.

Define $\mathbb I := \closedint 0 1$ as a closed real interval.

Let $\gamma : \mathbb I \to A$ be a loop in $A$ with base point $a$.

Let $\sigma : \mathbb I \to \set a$ be the constant function.

Constant Function is Continuous shows that $\sigma$ is continuous, so $\sigma$ is a loop with base point $a$.

Define $H: \mathbb I \times \mathbb I \to A$ by:


 * $\map H { s, t } = t \map {\gamma} s + \paren { 1-t } a$

By definition of star convex set, we have $\map H { s, t } \in A$ for all $\tuple { s, t } \in \mathbb I \times \mathbb I$.

Define $F : \mathbb I \times \mathbb I \to A \times \mathbb I$ by $ \map F { s, t } = \tuple { \map { \gamma }{ s }, t }$.

As a loop is continuous, it follows that $F$ is continuous.

Define $\tilde G : V \times \mathbb I \to V$ by $\map { \tilde G }{ v, t } = t v + \paren { 1-t } a$.

By definition of topological vector space, it follows that $\tilde G$ is continuous.

Let $G = \tilde G {\restriction_{\paren {A \times \mathbb I} \times A} }$ be the restriction of $\tilde G$ to $\paren {A \times \mathbb I} \times A$.

Restriction of Continuous Mapping is Continuous:Topological Spaces shows that $G$ is continuous.

Composite of Continuous Mappings is Continuous shows that $ H = G \circ F$ is continuous.

For all $s, t \in \mathbb I$, we check that $H$ is a path homotopy between $\gamma$ and $\sigma$:

It follows that all loops with base point $a$ are path-homotopic with $\sigma$.

Relative Homotopy is Equivalence Relation shows that all loops with base point $a$ are path-homotopic with each other.

This implies that the fundamental group $\map { \pi_1 } { A, a }$ is trivial.

Star Convex Set is Path-Connected shows that $A$ is path-connected.

Hence the result.