Consider the real-valued function $M:=\{(x,y,z)| (2 - (x^2 + y^2)^{1/2})^2 + z^2=1\}$ defined on $\mathbb{R}^3-\{(0, 0, z)\}$. Show that the manifold $N=\{(x,y,z)\in \mathbb{R}^3|x^2+y^2= 4\}$ is transverse to M. Identify the resulting manifold $N\cap M$.
My Attempt:
Pardon me for something vacuous, as I am a beginner in differential topology.
All I know is that it's intersection is union of two disjoint circles. It's clear to me that $M$ is the equation of a torus. Now, substituting the value $x^2+y^2=4$ in the equation $M$, we get two circles each $x^2+y^2=4$ one at $z=1$ and the other at $z=-1$. By definition of transversality,we should have $T_p(M)+T_{p}(N)=T_{p}\mathbb{R}^3\simeq\mathbb{R}^3,\forall p\in M\cap N$. But then I don't understand how the tangent spaces spans $\mathbb{R}^3$ being parallel(I guess)??
 A: To find whether or not intersection of these surfaces $T^2=\lbrace (x,y,z): (2-\sqrt{x^2+y^2})+z^2=1\rbrace$ and $C=\lbrace (x,y,z): x^2+y^2=4\rbrace$ is transverse, lets first look at the intersection. So 
\begin{align}T^2\cap C&=\lbrace\ (x,y,z): x^2+y^2=4\text{ and }(2-\sqrt{x^2+y^2})+z^2=1\rbrace\\
&=\lbrace\ (x,y,z): x^2+y^2=4\text{ and }(2-\sqrt{4})+z^2=1\rbrace\\
&=\lbrace\ (x,y,z): x^2+y^2=4\text{ and } z^2=1\rbrace
\end{align}
so they intersect in circles of radius 2 at heights $z=\pm 1$. Now suppose that $(x,y,1)$ is on the top circle and $x$ is positive. Then we can locally parametrize $T^2$ near $x$ with $\phi$ by solving for $z$ and choosing the positive branch. This gives $\phi:U\to\mathbb{R}^3$ by
$$\phi(x,y)=\left(x,y,\sqrt{1-(2-|r|)^2}\right)\quad (\text{where $r^2=x^2+y^2$})$$
Let $\psi:V\to\mathbb{R}^3$ be the local parametrization of $C$ around $(x,y,1)$ given by
$$\psi(x,z)=(x,\sqrt{4-x^2},z)$$
Then the tangent space of $T^2$ at $(x,y,1)$ is the image of $d\phi_{(x,y)}$ and the tangent space of $C$ at $(x,y,1)$ is the image of $d\psi_{(x,1)}$. These are
$$d\phi_{(x,y)}=\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\frac{x(2-|r|)}{|r|\sqrt{1-(2-|r|)^2}} & \frac{y(2-|r|)}{|r|\sqrt{1-(2-|r|)^2}} \\
\end{bmatrix}$$ and $$d\psi_{(x,1)}=\begin{bmatrix}
1 & 0 \\
\frac{-x}{\sqrt{1-x^2}} & 0 \\
0 & 1 \\
\end{bmatrix}$$
But remember that $x^2+y^2=4$ for $(x,y,1)$. That is, $r=2$. So the above $d\phi_{(x,y)}$ becomes
$$d\phi_{(x,y)}=\begin{bmatrix}
1 & 0 \\
0 & 1 \\
0 & 0\\
\end{bmatrix}$$ So the spanning set is $\lbrace d\phi_{(x,y)}(\vec{i}), d\phi_{(x,y)}(\vec{j}), d\psi_{(x,1)}(\vec{j})\rbrace=\lbrace\vec{i},\vec{j},\vec{k}\rbrace$
