# Show that a node of a real curve is homeomorphic to a cross

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.

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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 $t \mapsto (t,-t\sqrt{x+1})$. – Yang Zhou 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
Also, $C$ is a manifold, it is just not a submanifold of $\mathbb{R}^2$. – William Oct 21 '12 at 4:14