# Local intersection of a tubular neighborhood with a fixed tangent space

Let $$M\subset\mathbb{R}^n$$ be a smooth submanifold of dimension $$m, and a tubular neighborhood $$\mathcal{V}=E(\{(x,v)\in NM:|v|<\delta(x)\})$$ (diffeomorphic image of the smooth map $$E:NM\to\mathbb{R}^n$$ defined by $$E(x,v)=x+v$$) with $$\delta:M\to]0,\infty[$$ continuous function.

Fixed a point $$p$$, we can consider a local parametrization $$\varphi:\mathbb{R}^m\to U_p\subset M$$ with $$\varphi(0)=p$$ such that $$T_pM=\varphi'(0)(\mathbb{R}^m)$$.

From this, it can be shown that there exists a nbd $$B_r(0)\subset\mathbb{R}^m$$ such that the part $$p+\varphi'(0)(B_r(0))$$ of the affine tangent space $$p+T_pM$$ is contained in the tubular nbd $$\mathcal{V}$$, and there exists other nbd $$V_p\subset U_p$$ such that the affine normal spaces $$x+N_xM$$ intersects $$p+\varphi'(0)(B_r(0))$$ in a unique point $$p_x$$ for all $$x\in V_p$$ (show the image below in the case $$n=3$$ and $$m=1$$). My question:

Taking into account that the normal bundle $$NM$$ is a smooth submanifold of $$\mathbb{R}^{2n}$$, it can be proven that the map $$x\mapsto p_x$$ (well defined in $$V_p$$) is smooth too?

• Your picture really is in $\Bbb R^2$. You should be drawing $V$ as a solid tube (with circular cross-sections) if you're in $\Bbb R^3$. Also, having $V$ already, writing $V_p$ for a subset of $U_p$ is beyond confusing. When $p=x$ you're intersecting orthogonal affine subspaces; for $x$ near $p$, they are almost orthogonal and hence have a unique intersection. Can you set this up as an application of the implicit function theorem to prove smoothness? Aug 21, 2022 at 20:37
• Thanks for the hints, I was able to solve it locally in two ways, one was using the implicit function theorem and other was using transversality theory. I will respond twice to the post with images, showing the two ways. I would appreciate if you can check them. Aug 22, 2022 at 16:18
• I find your written solutions difficult to read. Something seems slightly fishy with the second, but it might be more helpful to write it without component-by-component notation. Your approach is, however, correct. The first seems right, but I wouldn't call this transversality theory. Knowing that the preimage of a regular value is a submanifold is elementary and doesn't even begin to use Sard's Theorem. Aug 22, 2022 at 21:14
• Thanks Ted, I'll have it in mind. Aug 24, 2022 at 0:02  