# Hyperplane such that intersects transversality

Let $$M$$ be a submanifold of $$\mathbb{R}^n$$ show that exists a hyperplane $$H$$ such that intersects transversaly t $$M$$

First I define $$F:M \times \mathbb{S}^{n-11} \rightarrow \mathbb{R}$$ given by $$F(x,y)= \langle x | y \rangle$$ then $$F$$ is a submersion because $$F$$ is the restriction to the submanifold $$M \times \mathbb{S}^{n-1}$$ of the submersion $$G: \mathbb{R}^{2n} \rightarrow \mathbb{R}$$ with $$G$$ the inner product in $$\mathbb{R}^n$$ then i try to applicated the transversality theorem but I not sure how, because these theorem only gives me to $$F_y \pitchfork Z$$ f.a.e $$y \in \mathbb{R}^{n}$$ and $$Z$$ any submanifold of $$\mathbb{R}$$ i think to the hyperplane what I am looking for is the inverse image $$F^{-1}_{y}(c)$$ with $$c$$ a regular value

Edit: I proved that $$0$$ is a regular value of $$F_y$$ and how $$\{0\}$$ is a submanifold of $$\mathbb{R}$$ then $$F_y \pitchfork \{0\}$$, now $$F_y^{-1}(0)$$ is the set such that the hyperplane $$H=\{w \in \mathbb{R}^n ; \langle w,v \rangle =0\}$$ intersects to $$M$$ can i conclude that these intersection is transversaly?

Any hint or suggestions I appreciated

We can suppose the origin does not belong to $$M$$. A hiperplane $$H\subset\mathbb R^n$$ is generated by $$n-1$$ independent vectors $$u_i$$, so we consider the mapping $$F:\varOmega\times(\mathbb R^{n-1}\setminus\{0\})\to\mathbb R^n:(u,t)\mapsto\sum_it_iu_i$$ Here $$u=(u_i), u_i\in\mathbb R^n$$, $$t=(t_i),t_i\in\mathbb R$$ and $$\varOmega\subset\mathbb R^{n\times(n-1)}$$ is the open set defined by rank$$(u)=n-1$$. Thus for each $$u=(u_i)$$ fixed, the image of the partial function $$F_u$$ is the hyperplane $$H_u$$ generated by the $$u_i$$'s, except the origin (because we excluded $$t=0$$). This mapping $$F$$ is a submersion. Indeed, fix any $$(u^0,t^0)$$ with say $$t^0_i\ne0$$, and for $$v\in\mathbb R^n$$ denote $$u=(0,\dots,v,\dots,0)$$. Then $$d_{(u^0,t^0)}F(u,0)=t_iv,$$ which shows $$d_{(u^0,t^0)}F$$ is onto. Thus, since $$F$$ is a submersion, it is transversal to $$M$$. Consequently, by density of transversality (the basic parametrized version which follows from the Sard-Brown theorem), for almost every $$u$$, the partial function $$F_u$$ is transversal to $$M$$. This means that the hyperplane $$H_u$$ is transversal to $$M$$ at every point except the origin, but this point does not belong to $$M$$. We are done.