# Transversality of a mapping

The question I have is:

Show that the mapping $g:R^2 \rightarrow R^3$ given by : $y_1 = x_1 (x_1 ^2 -x_2 ^2 +1), y_2=x_2, y_3=x_1 ^2$ is transversal to all lines $y_2 = \textrm{constant}$ in the plane $y_1=0$ except for two of them. Specify these exceptional lines.

Now, I thought that since the tangent space of $y_1 = \textrm{constant}$ at any point was spanned by (1,0,0), (0,0,1), then transversality would fail where the rank of the matrix of the Jacobian of g augmented with the columns (1, 0, 0) and (0, 0, 1) is less than 3 (the dimension of the tangent space of the source).

However, I'm looking at the 3x3 minors of this matrix (they should all equal zero for the rank to be less than 3) and one of the 3x3 minors always seems to be -1. This would mean that the function is always transversal to lines $y_2 = \textrm{constant}$, which (looking at how the question in phrased) doesn't seem to be right.

I think that I must be calculating the minors wrong. The minor I'm having trouble with has rows: $(-2x_1 x_2, 1, 0), (1,0,0), (0,0,1)$ which, expanding along the bottom row has determinant -1.

Could someone please tell me what I'm doing wrong?

Many thanks in advance. (Note: I'm not looking for a complete answer to the question, rather for someone to tell me what I'm doing wrong in my solution so far).

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The tangent space of a line $y_2=\textrm{constant}$ in the plane $y_1=0$ is spanned only by $(0,0,1)$ (tangent spaces to lines are 1-dimensional!)
The $4 \times 3$ matrix you've been looking at would be what you wanted if you were looking at whether your map was transverse to the planes $y_2=\textrm{constant}$.