# 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 $$\forall z \in G(f) \cap H,\, T_z (M\times N) = T_z G(f) + T_z H$$

• 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. Dec 11, 2022 at 18:20
• @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? Dec 12, 2022 at 0:53
• 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. Dec 13, 2022 at 4:53

$$(\implies)$$ 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.