Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does anyone know a reference to the classifications of local homeomorphisms from a closed line segment into $\mathbb{R^2}$ ? I suspect it is given by the minimal number of self intersections of the image curve.

share|cite|improve this question
Classification up to a global homeomorphism of $\mathbb R^2$? – user31373 Jun 7 '12 at 15:57
If the map is given by f:[a,b]\to R^2, I am interested in a classification up to homotopy (in R^2) with endpoints f(a) and f(b) fixed. – Dan Gallo Jun 7 '12 at 16:35
Then my guess is that you can homotope anything into the straight line segment, though preserving the local homeomorphism property along the way will give a bit of headache. – user31373 Jun 7 '12 at 16:38
Thanks. You are correct. I've got to rethink my question. – Dan Gallo Jun 7 '12 at 17:32

Fact. Every local homeomorphism of $[0,1]$ into $\mathbb R^2$ is homotopic (through local homeomorphisms, without moving the endpoints) to a polygonal path.

Once the fact is proved, one can reduce the number of intermediate vertices by homotoping the broken line $a_{i-1}a_ia_{i+1}$ to $a_{i-1}a_{i+1}$. When the intermediate vertices are gone, we have a line segment. Hence, the proposed classification has only one class.

Proof of the fact: By partitioning $[0,1]$, the problem reduces to the case of a homeomorphism $f:[0,1]\to\mathbb R^2$. Let $\Phi$ be a conformal map of $\mathbb C\setminus [0,1]$ onto $\mathbb C\setminus f([0,1])$ with boundary values $\Phi(0)=f(0)$ and $\Phi(1)=f(1)$. Let $\gamma_\kappa$, $0<\kappa<1$, be the circular arc with endpoints $0$ and $1$ and center $1/2-i/\kappa$. The curves $\Phi\circ \gamma_\kappa$, after suitable reparametrization, form a homotopy between $f$ and a smooth embedding of $ [0,1]$ into the plane. A smooth embedding can be approximated by a polygonal curve, to which it is related by a straight-line homotopy. $\Box$

(There may be an easier proof of the fact, but it escapes me now.)

share|cite|improve this answer
I don't follow. Why does it suffice to work with a homeomorphism? – Grumpy Parsnip Jun 29 '13 at 4:29
@GrumpyParsnip Because when $f:[0,1]\to\mathbb R^2$ is a local homeomorphism, one can partition $[0,1]$ into subintervals on each of which $f$ is a homeomorphism. The homotopy to polygonal path fixes the endpoints. The concatenation of polygonal paths is also a polygonal path. (OK, one has to make sure that after concatenation we don't get overlap of successive segments, but this holds generically, e.g., if all of them have different slopes). – ˈjuː.zɚ79365 Jun 29 '13 at 4:34
Okay, I get it now. Nice argument! – Grumpy Parsnip Jun 29 '13 at 4:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.