A set S is called star-like if there exists a point $\alpha\in S$ such that the line segment connecting $\alpha$ and z is contained in S for all $z\in S$. Show that a star-like region is simply connected.
My answer
Show that
$γ:γ(t)=tz+(1−t)α, t≥1$ is contained in the complement for any z in the complement
Let $\gamma$ represents the portion of the ray from $\alpha$ through z to $\infty$, starting at z. Thus, if z is in the complement of S, so is all of $\gamma$. For, if any $z_{1}\in \gamma$ belonged to S, so would the entire segment connecting $\alpha$ and $z_{1}$, including z.
Could anyone help to formalize the answer?
Best Answer
Here's a stronger algebraic-topological argument. We can show that a star-shaped set $S$ is not just simply connected, it is contractible. There is an $\alpha$ so for each $p\in S$ the line segment $\gamma_p(t) = tp + (1-t)\alpha$ has $\gamma_p\in S$ for all $t$. Now define $H(p,t):S\times [0,1]\to S$ by $(p,t)\mapsto \gamma_p(t)$. This is a strong deformation retract of $S$ onto $\{\alpha\}$, so $S$ has the homotopy type of a point.
For every topological space $X$, every map $f:X\to S$ is homotopic to the constant map $X\to\{\alpha\}$. The intuition here is to just "slide" the map to $\alpha$, more precisely via the homotopy $F(x,t) = H(f(x),t)$. So in particular we can see that any map $\phi:\mathbb{S}^1\to S$ can be contracted to the constant map $\phi:\mathbb{S}^1\to \{\alpha\}$, hence $\pi_1S = 1$.