Transversality in $\mathbb{R}^2$

Question

Define, for some $a\in\mathbb{R}, f_{a}:\mathbb{R}\to\mathbb{R}^{2}$ by $f_{a}(p)=(p,a)$ and consider $N\subset\mathbb{R}^{2}, \ N=\{(x,x^2); x\in\mathbb{R}\}.$ Analysis of the transversality of $f_{a}$ with $N.$ Could someone help me, please?

Geometrically, we can see that for $a\leq 0,$ $f_{a}$ isn't transversality to $N$. How to prove it?

So, $d(f_{a})_{p}(v)=(v,0)$ and $T_{(p,p^2)}N=\{(v,2pv)| v \in \mathbb{R}\}.$

How and why does $ a $ influence this?

Answer 1

Let me recall that:

Given $X$ and $Y$ two smooth manifolds without boundary, a smooth map $f\colon X\to Y$, and $Z$ a submanifold of $Y$, then $f$ is transversal to $Z$ whenever for all $x\in f^{-1}(Z)$, $T_xf(T_xX)+T_{f(x)}Z=T_{f(x)}Y$ holds true.

Therefore, whenever $f^{-1}(Z)$ is empty, $f$ is automatically transverse to $Z$. In particular, if $a<0$, then ${f_a}^{-1}(N)$ is empty and $f_a$ is transverse to $N$.

Furthermore, for all $p\in\mathbf{R}$ and all $v\in\mathbf{R}$, $T_pf_a(v)=(v,0)$, indeed, carry the following computations: \begin{align*} f_a(p+v)&=(p+v,a),\\ &=(p,a)+(v,0),\\ &=f_p(a)+(v,0)+O(\|v\|^2), \end{align*} hence as $v\mapsto (v,0)$ is linear, it is $T_pf_a(v)$.

From there, distinguish the three following cases:

• If $a<0$, then ${f_a}^{-1}(N)$ is empty and $f_a$ and $N$ are transverse.
• If $a=0$, then ${f_a}^{-1}(N)=\{0\}$, $T_0f_a(T_0\mathbf{R})=\{(v,0);v\in\mathbf{R}\}$ and $T_{(0,0)}N=\{(v,0);v\in\mathbf{R}\}$, therefore any point $(x,y)\in T_{(0,0)}\mathbf{R}^2=\mathbf{R}^2$, with $y\neq 0$, cannot be written as a sum of elements in $T_0f_a(T_0\mathbf{R})$ and $T_{(0,0)}N$, meaning that $f_a$ and $N$ are not transverse.
• If $a>0$, then ${f_a}^{-1}(N)=\{-\sqrt{a},\sqrt{a}\}$, $T_{\pm\sqrt{a}}f_a(T_0\mathbf{R})=\{(v,0);v\in\mathbf{R}\}$, $T_{(\pm\sqrt{a},a)}N=\{(v,\pm 2\sqrt{a}v);v\in\mathbf{R}\}$, therefore given any $(x,y)\in T_{(\pm\sqrt{a},a)}\mathbf{R}^2=\mathbf{R}^2$, it holds: $$(x,y)=\left(x\mp\frac{y}{2\sqrt{a}},0\right)+\left(\pm\frac{y}{2\sqrt{a}},y\right),$$ which is a sum of an element of $T_{\pm\sqrt{a}}f_a(T_0\mathbf{R})$ and $T_{(\pm\sqrt{a},a)}N$, meaning that $f_a$ and $N$ are transverse.

Geometrically, you are asked to determine when the line $y=a$ intersects transversely the parabola $y=x^2$, as these are two one-dimensional submanifolds of $\mathbf{R}^2$, which is a two-dimensional manifold, this happens whenever $y=a$ is not tangent to $y=x^2$ (as two vectors of $\mathbf{R}^2$ span $\mathbf{R}^2$ whenever they are not colinear), this explains the three cases above.