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What is the definition of the curve?

a. Is the image of $x^2+y^2=1(x\neq0)$ a curve?

b. Is a point a curve?

Here are the definitions I found in Wikipedia that may help.

  1. A curve is a topological space which is locally homeomorphic to a line.

  2. Let I be an interval of real numbers (i.e. a non-empty connected subset of $R$). Then a curve is a continuous mapping $\gamma:I\to{}X$, where X is a topological space.

Two objects are homeomorphic if they can be deformed into each other by a continuous, invertible mapping. Right? But the second definition doesn't mention about the invertible.

Where can I get a rigorous definition of curve? Or in which topology textbook can I find the answer?

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    @J.M. then can you get a continuous, invertible mapping?2011-12-02

1 Answers 1

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I think the second definition that wiki gave is the most intuitively. Namely, one can think of a curve in a topological space $X$ as being a "path" traced out continuously, where the trace is indexed by time. From this it immediately follows that the a) is not a curve. One can't take a stroll in $\mathbb{R}^2$ and hope to jump from the above the $x$-axis to below. Similarly, a point is a curve. It corresponds to the path where we go nowhere. This is automatically continuous since, for us to screw up continuity (make a jump), we'd have to go somewhere!

More mathematically, a) cannot be a curve since any curve is the continuous image of the connected space $I=[0,1]$ and thus itself is connected. Since a) is not connected, it can't be a curve. b) is a curve since any constant map is continuous.

Perhaps intriguing though to you is the notion of $1$-manifolds which correspond to spaces that "locally look like (non-trivial) curves" (your definition 1.). In this case a) is a $1$-manifold since near any point a) looks like a bending/scaling of $I$ whereas b) is not since you can't scale $I$ down to a point! More rigorously, you can take the upper and lower semi-circles and paramaterize them by the maps $t\mapsto e^{\pi i t}$ for $t\in(0,1)$ and $t\mapsto e^{-\pi i t}$ $t\in(0,1)$.

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    @AlexYoucis I'm not familiar with geometry terms about exponential map and fundamental period. But I know that if the mapping is invertible, it should be bijective. Right? Can you recommend some geometry textbook? Thanks.2011-12-03