I've come across a question that seems to make sense on an intuitive level, but I'm lacking for any sort of concrete evidence.

I'm to show that for a given base point (call it $\alpha$) in an open subset of $\mathbb{C}$, every closed (especially continuous) loop/path from $\alpha$ to $\alpha$ is homotopic to a closed, piecewise smooth path about the same point $\alpha$ (for example a closed polygonal chain).

Again it seems rather conceptually intuitive but I'm failing to produce anything concrete towards this claim. I've seen that such closed, piecewise smooth paths/loops are homotopic to paths of the form $\ e^{2\pi itk}$ for some $(k\,\epsilon \,\mathbb{Z})$, which would be a continuous closed path, but I wouldn't know how to further generalize for the above basepoint/subset or if this is even a valid direction to head in.

Thoughts/starting point?


Let $\gamma:[0,1] \to \Omega$ be your closed continuous path. For convenience, consider it as a $1$-periodic function $\gamma: \mathbb R \to \Omega$.

Consider a bump function $\eta: \mathbb R \to [0,1]$, i.e. $\eta$ is $C^\infty$, compactly supported and $\int_{\mathbb R} \eta=1$. Define the dilations $\eta_\epsilon(x) = \frac1{\epsilon} \eta\bigl(\frac x \epsilon\bigr)$. A standard result now is that the convolution $\gamma^\epsilon = \eta_\epsilon * \gamma$ defined by $(\eta_\epsilon * \gamma)(t) = \int_{\mathbb R} \eta_\epsilon(\tau) \gamma(t-\tau) \, dt$ is $C^\infty$ and $\gamma^\epsilon \to \gamma$ uniformly in $[0,1]$.

In particular, there is an $\epsilon>0$ such that the line segment connecting $\gamma(t)$ and $\gamma^\epsilon(t)$ is contained in $\Omega$, for all $t \in [0,1]$. This is because the image of $\gamma$ is compact, hence a small tubular neighborhood of it is contained in $\Omega$. This now implies that $\gamma$ and $\gamma^\epsilon$ are homotopic, $\gamma^\epsilon$ being smooth.

Of course, this does not fix the endpoint. To deal with this, you can simply translate $\gamma^\epsilon$ within $\Omega$ by the right vector, assuming that $\epsilon$ is small enough.


Here's a sketch of how I would prove it, though as pointed out in the comments it's probably harder than this:

  • Any closed path is compact, since it is the image of a compact set ($S^1$) under a continuous map.

  • Imagine now covering the path with open balls. This forms an open cover of the path. Since the path is compact, there exists a finite subcover of the path.

  • Within each open ball in this finite subcover, the path can be homotopically deformed to a straight line. (Care will have to be taken in the overlap regions to ensure that this homotopy is well-defined in the intersections between the open sets.)

This then defines a polygonal path that is homotopic to the original path.

  • $\begingroup$ Can you say how to take care in the overlap regions? The path can be quite ugly, e.g. it can enter each overlap countably many times. $\endgroup$ – user296355 Jun 26 '17 at 14:40
  • $\begingroup$ @cbilz: Good point. I had implicitly been assuming that the intersection of the path with each overlap region has a finite number of connected components. $\endgroup$ – Michael Seifert Jun 26 '17 at 15:56

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