Is This Enough To Show That Two Subsets Of The Plane Are Not Diffeomorphic?

Introduction

In the realm of differential topology, the concept of diffeomorphism plays a crucial role in understanding the topological properties of manifolds. Two subsets of a manifold are said to be diffeomorphic if there exists a smooth map between them that has a smooth inverse. In this article, we will explore the question of whether a given map is sufficient to show that two subsets of the plane are not diffeomorphic. Specifically, we will examine the case of $\mathbb{R}^2$ and $\mathbb{R}^2 - \{(0,0)\}$.

Background

To approach this problem, we need to understand the basics of differential topology and the concept of diffeomorphism. A diffeomorphism is a smooth map between two manifolds that has a smooth inverse. In other words, it is a map that preserves the smooth structure of the manifolds. The existence of a diffeomorphism between two subsets of a manifold implies that they are topologically equivalent.

The Map $F(x,y) = (x,y)/(x^2+y^2)$

The map $F(x,y) = (x,y)/(x^2+y^2)$ is a smooth map from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$. This map is defined as long as $x^2+y^2 \neq 0$, which is true for all points except the origin. The map $F$ can be thought of as a stereographic projection from the origin to the rest of the plane.

Properties of the Map $F$

The map $F$ has several important properties that make it useful for our purposes. First, it is a smooth map, meaning that it has continuous partial derivatives of all orders. Second, it is a local diffeomorphism, meaning that it is a diffeomorphism on a neighborhood of each point. Finally, it is a proper map, meaning that it is a closed map and its preimage is compact.

The Preimage of the Origin

The preimage of the origin under the map $F$ is the set of all points $(x,y)$ such that $F(x,y) = (0,0)$. This set is precisely the set of all points on the unit circle centered at the origin. In other words, the preimage of the origin is the unit circle.

The Preimage of a Neighborhood of the Origin

The preimage of a neighborhood of the origin under the map $F$ is the set of all points $(x,y)$ such that $F(x,y)$ is in the neighborhood. This set is a neighborhood of the unit circle.

The Map $F$ is Not a Diffeomorphism

The map $F$ is not a diffeomorphism from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$. To see this, consider the preimage of a neighborhood of the origin. This preimage is a neighborhood of the unit circle, but it is not a neighborhood of the origin. In other words, the preimage of a neighborhood of the origin is not a neighborhood of the origin.

Conclusion

In conclusion, the map $F(x,y) = (x,y)/(x^2+y^2)$ is not a diffeomorphism from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$. This is because the preimage of a neighborhood of the origin is not a neighborhood of the origin. Therefore, we have shown that $\mathbb{R}^2$ and $\mathbb{R}^2 - \{(0,0)\}$ are not diffeomorphic.

Further Discussion

The result we have obtained is a special case of a more general result in differential topology. Specifically, it is a consequence of the fact that a manifold is not diffeomorphic to its complement if it has a non-trivial fundamental group. In the case of $\mathbb{R}^2$, the fundamental group is trivial, but the result still holds.

References

• [1] de Rham, G. (1955). Variétés différentiables. Hermann.
• [2] Hirsch, M. W. (1976). Differential topology. Springer-Verlag.
• [3] Milnor, J. W. (1963). Morse theory. Princeton University Press.

Appendix

In this appendix, we provide a proof of the fact that the map $F(x,y) = (x,y)/(x^2+y^2)$ is not a diffeomorphism from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$. The proof is based on the fact that the preimage of a neighborhood of the origin is not a neighborhood of the origin.

Proof

Suppose that the map $F$ is a diffeomorphism from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$. Then, the preimage of a neighborhood of the origin is a neighborhood of the origin. But this is not true, since the preimage of a neighborhood of the origin is a neighborhood of the unit circle, not the origin. Therefore, the map $F$ is not a diffeomorphism from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$.

Final Thoughts

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Q: What is the main result of this article?

A: The main result of this article is that the map $F(x,y) = (x,y)/(x^2+y^2)$ is not a diffeomorphism from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$. This implies that $\mathbb{R}^2$ and $\mathbb{R}^2 - \{(0,0)\}$ are not diffeomorphic.

Q: What is the significance of this result?

A: This result has important implications for the study of differential topology and the concept of diffeomorphism. It shows that two subsets of the plane can be topologically equivalent but not diffeomorphic.

Q: What is the relationship between diffeomorphism and topological equivalence?

A: Diffeomorphism and topological equivalence are related but distinct concepts. Two subsets of a manifold are topologically equivalent if they can be transformed into each other by a continuous map. Two subsets of a manifold are diffeomorphic if they can be transformed into each other by a smooth map.

Q: What is the difference between a smooth map and a continuous map?

A: A smooth map is a map that has continuous partial derivatives of all orders. A continuous map is a map that preserves the topological properties of the manifold.

Q: Can you provide an example of a smooth map that is not a diffeomorphism?

A: Yes, the map $F(x,y) = (x,y)/(x^2+y^2)$ is a smooth map that is not a diffeomorphism from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$.

Q: What is the preimage of a neighborhood of the origin under the map $F$?

A: The preimage of a neighborhood of the origin under the map $F$ is a neighborhood of the unit circle.

Q: Why is the preimage of a neighborhood of the origin not a neighborhood of the origin?

A: The preimage of a neighborhood of the origin is not a neighborhood of the origin because it is a neighborhood of the unit circle, not the origin.

Q: What is the relationship between the preimage of a neighborhood of the origin and the concept of diffeomorphism?

A: The preimage of a neighborhood of the origin is related to the concept of diffeomorphism because it shows that the map $F$ is not a diffeomorphism from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$.

Q: Can you provide a proof of the fact that the map $F$ is not a diffeomorphism?

A: Yes, the proof is based on the fact that the preimage of a neighborhood of the origin is not a neighborhood of the origin.

Q: What are the implications of this result for the study of differential topology and the concept of diffeomorphism?

A: This result has important implications for the study of differential topology and the concept of diffeomorphism. It shows that two subsets of the plane can be topologically equivalent but not diffeomorphic.

Q: Can you provide a reference for this article?

A: Yes, the references for this article are:

• [1] de Rham, G. (1955). Variétés différentiables. Hermann.
• [2] Hirsch, M. W. (1976). Differential topology. Springer-Verlag.
• [3] Milnor, J. W. (1963). Morse theory. Princeton University Press.

Q: What is the final thought of this article?

A: In conclusion, we have shown that the map $F(x,y) = (x,y)/(x^2+y^2)$ is not a diffeomorphism from $\mathbb{R}^2$ to $\mathbb{R}^2 - \{(0,0)\}$. This result has important implications for the study of differential topology and the concept of diffeomorphism. We hope that this article has provided a useful introduction to this topic and has inspired further research in this area.