Paths as homeomorphisms of the closed interval from 0 to 1
Previously, I defined a path in terms of a continuous function from a closed interval on the reals to a set of points in a topological space.
Because the function is continuous, by definition, the resultant image is homeomorphic to the closed interval on the reals. Because any closed interval on the reals is itself homeomorphic to the specific closed interval [0, 1] then the image of a path can be said to be homeomorphic to the real interval [0, 1].
UPDATE (2005-06-01): As Michael Hudson points out in a comment, a path will only be homeomorphic to the closed interval [0, 1] if it doesn't cross over itself. Homeomorphisms require the function to be bijective, continuous and have a continuous inverse. A path that crosses over itself doesn't meet these criteria.
UPDATE: next post
Comments (5)
James Tauber on June 1, 2005:
Yes, you're right. The inverse has to be continuous for homeomorphisms.
I should have clarified that the path can't cross over itself.
Michael Hudson on June 2, 2005:
For added contortions you could think about a map like "x |-> sin(10x)" which is a map from [0,1] to [-1,1], which spaces are homeomorphic, but the map is far from being a homeomorphism...
James Tauber on June 3, 2005:
In the case of your sin example, would it make a difference depending on whether you defined the path to be the map versus the image?
If the "path" is just the image then the path defined by x |-> sin(10x) is the closed interval [-1, 1] which *is* homeomorphic to the interval [0, 1], right?
David Feldman on Oct. 28, 2005:
The fact that the image is homeomorphic (which it clearly is) should perhaps be distinguished from the question of whether the map establishes that it is homeomorphic. In the case of x |-> sin(10x) the image of the map is homeomorphic to the domain ... but you wouldn't guess by looking at the map you used. In this case, you would need to look at another map like x | 2x-1 to show that the two intervals were homeomorphic. Perhaps that is what you were saying though.
DF
Last Modified: June 1, 2005
Author: James Tauber
Michael Hudson on May 31, 2005:
Huh? If the image self-intersects, I don't think you have a continuous inverse.