I am working on a practice question for differential geometry and reviewing the provided solutions - but I am having a bit of trouble connecting the ideas of transverse planes and regular curves,
The question is this:
Prove that if all normal lines of a connected surface $\sum$ meet a fixed straight line, then every point of $\sum$ has a neighbourhood that is contained in a surface of revolution.
The solution proceeds as follows:
Let $r$ be a fixed line which is met by the normals of the surface $\sum$. Consider the plane $Q$ which is perpendicular to the line $r$ and contains a point $p \in \sum$.
Here is where my confusion is, the solution then says:
Since the normal at $p$ meets the line $r$, it must be that $Q$ is transverse to the tangent plane at $p$. Thus, in a neighbourhood of $p$, $Q \cap \sum$ is regular curve.
Looking up the definition of transverse, it says that smooth manifolds intersect transversally if at every point of intersection, their separate tangent spaces at that point together generate the tangent space of the ambient manifold at that point.
I'm having a bit of trouble with this definition - if $Q$ and $T_p \sum$ are transverse, does this mean that the tangent vectors of $p$ in $Q$ and $T_p \sum$ generate the tangent space at $p$? (Would this be the whole space?) - what is the 'ambient manifold' in this case? Is it the curve where $Q$ and $T_p \sum$ intersect?
Why does this then mean that $Q \cap \sum$ is a regular curve?
Thanks for any help you can give me.