Every "star-shaped" set is simply connected

This is based off of a question in Serge Lang's Complex Analysis book, though much harder than the version of the question in the text.

Call a set $S\subseteq\Bbb{C}$ star-shaped if there exists a point $z_0\in S$ such that the line segment between $z_0$ and any point $z\in S$ is contained in $S$. Prove that a star-shaped set is simply connected, that is, every closed path $\gamma$ is homotopic to a point, where a path is defined as any continuous map $\gamma : [0,1]\to\Bbb{C}$.

The version in the text assumed that $S$ is open, which makes the problem many times easier, and assumed that $\gamma$ was continuously differentiable except at finitely many points.

So far, I've eliminated the case where $z_0$ is in the image of $\gamma$, converted to polar coordinates around $z_0$, managed to isolate $\gamma$ to an arbitrarily sliver of the plane, and I would like to make it homotopic to a line from $\gamma(0)$ to $z_0$, but I can't really see how to do it.

Bonus points if you can prove it for the general $\Bbb{R}^n$ case.

MAJOR EDIT: I'm looking for a homotopy that preserves the start and end points.

Solution 1:

Here's a homotopy from $\gamma$ to a constant at $x_0$:

$$ H(x,t) = (1-t) \gamma(x) + t x_0 $$

At $t = 0$, this gives $\gamma$; at $t = 1$, ti gives a constant path at $x_0$.

Thus any two paths are homotopic (because both are homotopic to the constant path, by a homotopy analogous to $H$). Hence the start-shaped set is simply connected.

The only subtle points are

• Is the image of $H$ entirely within the set $S$? (Yes, by definition of star-shaped) '
• For each fixed $t$, is $x \mapsto H(x, t)$ a smooth curve for $0 \le x \le 1$? (Yes, because its tangent curve is just $ x \mapsto t\gamma'(x)$, by the chain rule and product rule.

Added post-comments:

You've asked for a homotopy that preserves endpoints, so here goes. First, let's say that $x_1 = \gamma(0)$. Second, I'm going to use $t$ as the parameter for the path, and $s$ as the parameter for the homotopy, because that's what I'm used to.

Let $$ H_0(t, s) = \begin{cases} x_1 & 0 \le t \le \frac{s}{3} \\ \gamma(\frac{t-\frac{s}{3}}{1 - \frac{2s}{3}}) & \frac{s}{3} \le t \le 1 - \frac{s}{3} \\ x_1 & \frac{s}{3} \le t \le 1 \end{cases} $$

Then $H_0$ is a homotopy from $\gamma$ to "$\gamma$-traversed-three-times-as-fast-with-a--pause-before-and-after-traversal" Let's call that new curve $\gamma_1$, so $$ \gamma_1(t) = H_0(t, 1). $$

Now let

$$ H_2(t, s) = \begin{cases} (1-s) x_1 + s ((1-3t)x_1 + 3t x_0)) & 0 \le t \le \frac{1}{3} \\ (1-s) \gamma_1(t) + s x_0 & \frac{1}{3} \le t \le \frac{2}{3} \\ s x_1 + (1-s) ((3-3t)x_0 + (3t-2) x_1))& \frac{2}{3} \le t \le 1 \end{cases} $$

That's a homotopy from $\gamma_1$ to a curve that goes from $x_1$ to $x_0$ in the first third, then sits at $x_0$, then returns to $x_1$ during the last third. Let's call that new curve $\gamma_2$.

Finally, let

$$ H_2(x, t) = (1 - t) \gamma_2(x) + tx_1. $$

That's a homotopy from $\gamma_2$ to the constant path at $x_1$.

We sequence these to get a homotopy from $\gamma$ to the constant path at $x_1$, and the homotopy, as you'll observe, is endpoint preserving, i.e., it's a homotopy of loops, not just paths.