Show that a star-like region is simply connected. 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?
 A: 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 $\pi_1S = 1$.
A: If I recall right (see this), you only have to show that your star set is path connected and that all such paths are homotopic, i.e. that one path can be deformed continuously into the other.
Here's one way to do it : Let $z_1, z_2 \in S$. Then the path connecting $z_1$ to $\alpha$ and then $\alpha$ to $z_2$, is a continuous path between $z_1$ and $z_2$. Given any two paths, the first one between $z_1$ and $z_2$, and the other one between $z_3$ and $z_4$, just deform your first path continuously into a point at $\alpha$, and then "expand it" in a path from $z_3$ to $z_4$, which gives you the homotopy you need.
Hope that helps,
A: I believe that the question should be "is every star shaped domain in the complex plane simply connected", and the answer is in the affirmative. However, not every simply connected domain is star shaped.
