Show that a star convex set $X \subset \mathbb{R^n}$ is simply connected.
I have the idea of the proof : There is a natural retraction of $X$ onto the set $\{x_0\}$, because every other point $x \in X$ can be continuously mapped by $f:X \to \{x_0\}$ to $x_0$.
( I am sure that I need to say : along the segment joining them, but seemingly I can write a retraction without specifying how to map $X$ onto $\{x_0\}$, it is strange !!! ).
the only open sets in $\{x_0\}$ are the empty set $\emptyset$ and the whole space $\{x_0\}$, preimage of both under $f$ is open in $X$. And since, obviously !, $f(x_0)=x_0$, $f$ is a retraction of $X$ onto $\{x_0\}$.
Since $\pi_1(\{x_0\})$ is trivial, the theorem implies that $X$ is simply connected.
My proof can not be correct, since I can find :D a retraction from a subset of $\mathbb{R^n}$ which is not star convex onto a point $x_0$ , and finally claim that $X$ is simply connected.
Any help is appreciated, Thanks !
A star-shaped domain is more than simply connected: It is contractible. For each $x$, there is a line segment $\gamma_x$ connecting $x$ to $x_0$. That is, $\gamma_x(0) = x$ and $\gamma_x(1) = x_0$. Can you use these paths to write a homotopy between the identity map and the constant map?