Regular value and transversal intersection of submanifolds I'm not sure how to solve the following exercise.
Let $M,N$ be smooth manifolds and $f:M\to N$ smooth. Show that $y\in N$ is a regular value of $f$ iff the submanifolds $G(f) = \{(x,f(x))\in M\times N : x\in M\}$ and $H=M\times\{y\}$ intersect transversally in $M\times N$
I know that $y\in N$ is regular if for all $x\in f^{-1}(y)\subset M$ it holds that $D_xf : T_xM \to T_yN$ is surjective and that the submanifolds intersect transverse if $\forall z \in G(f) \cap H,\, T_z (M\times N) = T_z G(f) + T_z H$
Thanks for your help.
 A: $(\implies)$ Suppose that $y\in N$ is a regular value. So, we have that for all
$x \in f^{-1}(y)$, $D_xf(T_x M) = T_{y}N$. Recall that we can identify
$T_{(x, y)}(M \times N)$ with $T_xM \oplus T_y N$. Observe that $G(f) \cap H =
f^{-1}(y)\times \{y\}$. Now, for any $x \in f^{-1}(y)$, we have
$$
T_{(x,y)}G(f) = \{(v, D_xf(v)) \in T_xM \oplus T_yN : v \in T_xM\}
$$
and,
$$
T_{(x,y)}H = \{(u, 0) : u \in T_xM\}.
$$
Now, let $(v', w') \in T_xM \oplus T_yN$. Since, $D_xf$ is surjective,
we may choose $v \in T_xM$ such that $D_xf(v) = w'$. So, we can write
$$
(v', w') = (v, D_xf(v)) + (v'-v, 0).
$$
So,
$$
T_{(x,y)}(M \times N) = T_xM \oplus T_yN = T_{(x,y)}G(f)  + T_{(x,y)}H.
$$
Therefore, $G(f)$ and $H$ are transverse to each other.
$(\impliedby)$  To prove the converse, use the decomposition
$$
T_{(x,y)}(M \times N) = T_{(x,y)}G(f)  + T_{(x,y)}H
$$
to find a preimage of $w \in T_yN$ under $D_xf: T_xM \to T_yN$ (basically, you need to reverse the steps above), which shows that $D_xf$ is surjective for all $x \in f^{-1}(y)$. So $y$ is a regular value.
