Null-homotopic loop in the complex plane

Question

In my complex analysis course, we started studying paths and loops in the complex plane and defined null-homotopy as follows:

A function $f$ with domain $U$, an open subset of $\mathbb{C}$, is said to be null-homotopic in $U$ if it is homotopic to a constant function.

But if we're talking about loops, doesn't that mean that the null-homotopic loop is just a point in $\mathbb{C}$?

Answer 1

Null-homotopy means that the loop can be "shrunk" to a point, it doesn't mean that the loop itself is a point. For instance, a loop $\gamma(t) = Re^{it}$ centered at the origin can be shrunk to a point in $\mathbb{C}$ by considering the homotopy

$$ H(t,s) \;\; =\;\; sRe^{it}. $$

When $s = 1$, we have our original loop $H(t,1) = \gamma(t)$, but when $s=0$, we have the constant function $H(t,0) = 0$.

If we changed the context of this to drawing a loop in the punctured plane $\mathbb{C}\setminus \{0\}$, then the mapping above would not serve as a homotopy to a constant function. In fact $\gamma(t)$ is not null-homotopic in $\mathbb{C}\setminus\{0\}$.