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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.

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    $\begingroup$ Use the fact that $T_{(x, f(x))} (M\times N) =T_xM\oplus T_{f(x)} N$ and try to figure out what $T_{(x, f(x))} G(f) $ and $T_{(x, y)} H$ are. $\endgroup$
    – J.V.Gaiter
    Dec 11, 2022 at 18:20
  • $\begingroup$ @J.V.Gaiter then one gets $T_{(x,f(x))}G(f)=T_x M \bigoplus T_{f(x)} f(M)$ and $T_{(x,y)}H=T_x M \bigoplus T_{y} \{y\}$. Therefore $\forall (x,z)\in G(f)\cap H$ (this implies $z=y \in f(x)$) it holds that $T_{(x,z)}G(f)+T_{(x,z)} H = T_x M \bigoplus T_{f(x)} f(M) + T_y \{y\}$. Is it right and where do I use that $y$ is regular? $\endgroup$ Dec 12, 2022 at 0:53
  • $\begingroup$ It is not true that $T_{(x,f(x))}G(f)=T_xM\oplus T_{f(x)}f(M)$ since $T_{f(x)}f(M)$ is not necessarily a well defined object, and even if it were $G(f)\neq M\times f(M)$ generically. $\endgroup$
    – J.V.Gaiter
    Dec 13, 2022 at 4:53

1 Answer 1

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$(\implies)$ 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.

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