Take the 2-minute tour ×
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.

This is related to this question.

Let $C = \{(x,y) \in \mathbb{R};\; x^3 + x^2 - y^2 = 0\}$ equipped with the subspace topology of the euclidian plane. I want to show that there's a neigbourhood $U$ of $0 \in C$ which is homeomorphic to a cross $X = (-1,1) \times \{0\} \cup \{0\} \times (-1,1)$.

This comes from an assignment to show that $C$ is no manifold. While visually clear, I want to do this as rigorously as possible.

share|improve this question
1  
The curve breaks into two analytic branches at the origin, you may want to parametrize the curve by $t \mapsto (t,t\sqrt{x+1})$ and $t \mapsto (t,-t\sqrt{x+1})$. –  Andrew Oct 20 '12 at 10:17
    
Andrew, I think you meant $t\mapsto (t, \pm t\sqrt{t + 1})$. Then of course the two branches (corresponding to $+$ and $-$) intersect transversally at the origin (i.e. the differential vectors of the curves at the origin are not colinear). Now consider the tangent lines through the origin to each curve, and project each curve (in the neighborhood of the origin) to the corresponding tangent line, and you're done. –  William Oct 21 '12 at 4:12
1  
Also, $C$ is a manifold, it is just not a submanifold of $\mathbb{R}^2$. –  William Oct 21 '12 at 4:14
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.