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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.

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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

1 Answer 1

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.)

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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

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