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How to show that the intersection of a plane and a smooth surface in 3D is "generically" a one-dimensional curve?

That is, if the intersection is not a curve and non-empty, then perturbing the plane a little bit, the intersection would become a curve.

I just tried to read Thom's transversality theorem but I don't really know how to apply.

It seems that by preimage theorem, if $f:\mathbb R^3\to\mathbb R^2$ is smooth, then $f^{-1}(y)$ is (generically?) a one dimensional manifold.

Let $X$ denote the plane and $Y$ denote the surface. This link" http://mathworld.wolfram.com/TransversalIntersection.html

says that if $X,Y$ intersect transversally, then $dim(X\cap Y)=2+2-3=1$. What theorem is this?

How to prove it is generic?

I just think the 2+2-3=1 from the Mathworld makes sense to beginners. I am just wondering what reference is it so I can continue to read it.

dodo
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  • @Conifold I appreciate you to point this out but I don't understand how to apply the generalization of pre-image theorem in that link. The link states that $dim(f^{-1})=dim Y- dim X$. In my case, which one is the curve and which one is the plane? – dodo Dec 19 '19 at 06:37
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    See the last paragraph in the accepted answer. Take as $f$ the inclusion map $i$ for $X$, since you want to perturb the plane make $X$ the plane. – Conifold Dec 19 '19 at 06:43
  • @Conifold I guess $Y$ is the surface, which the submanifold of $R^3$? Then the second theorem of the accepted answer states that the surface and $i^{-1}(Y)$ have the same dimensions? I don't get this because $Y$ is dimension 2 but $i^{-1}(Y)$ should be dimension 1. I am sure I don't understand it quite well, though. – dodo Dec 19 '19 at 07:04
  • No, it states that they have the same codimensions, in this case $1$. The surface $Y$ has codimension $1$ in $R^3$, as does $i^{-1}(Y)$ in the plane. Since the plane is 2D its dimension is also $1$. – Conifold Dec 19 '19 at 07:09

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