# Prove that it is a regular point

Good night, my friends, can you help me with these exercise?

Let $N$ a $k$-manifold, $X$ a compact $(k+1)$-manifold in $\mathbb R^N$, and $F:X\rightarrow N$ a differentiable map. Let $y\in \operatorname{Reg}(F)\cap \operatorname{Reg}(F|_{\partial X})$ and let $J$ a connected component of $F^{-1}$. Suppose $\partial J\neq\emptyset$, fix a diffeomorphism $c:[0,1]\rightarrow J$, choose a base $\{w_1,\ldots,w_k\}$ de $T_yN$ and let $t_0\in[0,1]$ arbitrary. Fixe $h:U\rightarrow h(U)$ a local parametrization of $X$ in $c(t_0)$ and denote by $\langle \cdot,\cdot \rangle$ the usual inner product of $\mathbb R^N$.

Consider $\Phi:c^{-1}(U)\times \mathbb R^{k+1}\rightarrow T_yN\times\mathbb R$ defined by

$$\Phi(t,u)=(d(F\circ h)_{h^{-1}\circ c(t)}\cdot u,\langle dh_{h^{-1}\circ c(t)},c'(t) \rangle).$$

Show that $(w,0)\in \operatorname{Reg}(\Phi)\cap \operatorname{Reg}(\Phi|_{\partial (c^{-1}(U)\times \mathbb R^{k+1})})$

for all $w\in T_yN$.

Let $(a,b)\in \Phi^{-1}(w,0)$, then I have to show that $D\Phi_{(a,b)}$ is surjective, my problem is finding $D\Phi_{(a,b)}$.

## Thank you

This is what I have done Let $p\in h^{-1}(h(U)\cap J)$, i.e., $h(p)=c(t)$, or $p=h^{-1}(c(t))$, onde $t\in[0,1]$, then if $\Phi=(\Phi_1,\ldots\Phi_k,\Phi_{k+1})$ from the definition of $\Phi$ we have that $$d(F\circ h)_p(u)=(\Phi_1(t,u),\ldots\Phi_k(t,u) )$$ and $$\left<{dh_p(u),c'(t)}\right>=\Phi_{k+1}(t,u).$$ Now, let $(V,\psi)$ be a parametrization of $F(c_{t_0})$, tal que $F(U)\subset V$. Then $$d(F\circ h)_p=d(g^{-1}\circ F\circ h)_p.$$

Let $\phi:=g^{-1}\circ F\circ h$, $u=(u_1,\ldots,u_{k+1})\in R^{k+1}$.

$$\begin{array}{l l l} d(F\circ h)_p(u)&=&\left[ \begin{array}{ccc} \cfrac{\partial \phi_1}{\partial x_1}(p) &\cdots& \cfrac{\partial \phi_1}{\partial x_{k+1}}(p) \\ \vdots & \dots & \vdots \\ \cfrac{\partial \phi_k}{\partial x_1}(p) &\cdots& \cfrac{\partial \phi_k}{\partial x_{k+1}}(p) \end{array} \right]\left[ \begin{array}{c} u_1\\ \vdots \\ \vdots \\ . \\ u_{k+1} \end{array} \right]\\ & &\\ &=&\left[ \begin{array}{c} \displaystyle\sum_{i=1}^{k+1}u_i\cfrac{\partial \phi_1}{\partial x_i}(p)\\ \vdots \\ \displaystyle\sum_{i=1}^{k+1}u_i\cfrac{\partial \phi_k}{\partial x_i}(p)\\ \end{array} \right]_{k\times{1}} \end{array}$$

By definition we have $$\cfrac{\partial F_l }{\partial x_i}(h(p))=\cfrac{\partial \phi_l}{\partial x_i}(p),l=1,\ldots,k.$$

Hence

$$d(F\circ h)_p(u)=\left[ \begin{array}{c} \displaystyle\sum_{i=1}^{k+1}u_i\cfrac{\partial F_1}{\partial x_i}(h(p))\\ \vdots \\ \displaystyle\sum_{i=1}^{k+1}u_i\cfrac{\partial F_k}{\partial x_i}(h(p))\\ \end{array} \right].$$

Next, let $(W,r)$ be another parametrization of $c(t_0)$, so $$dh_p=d(r^{-1}\circ h)_p.$$

Denoting $\psi=r^{-1}\circ h$,

$$\begin{array}{l l l} dh_p(u)&=&\left[ \begin{array}{ccc} \cfrac{\partial \psi_1}{\partial x_1}(p) &\cdots& \cfrac{\partial \psi_1}{\partial x_{k+1}}(p) \\ \vdots & \dots & \vdots \\ \cfrac{\partial \psi_{k+1}}{\partial x_1}(p) &\cdots& \cfrac{\partial \psi_{k+1}}{\partial x_{k+1}}(p) \end{array} \right]\left[ \begin{array}{c} u_1\\ \vdots \\ \vdots \\ . \\ u_{k+1} \end{array} \right]\\ & &\\ &=&\left[ \begin{array}{c} \displaystyle\sum_{i=1}^{k+1}u_i\cfrac{\partial \psi_1}{\partial x_i}(p)\\ \vdots \\ \displaystyle\sum_{i=1}^{k+1}u_i\cfrac{\partial \psi_{k+1}}{\partial x_i}(p)\\ \end{array} \right]_{(k+1)\times{1}} \end{array}$$

In the same way we get

$$\cfrac{\partial h_l }{\partial x_i}(p)=\cfrac{\partial \psi_l}{\partial x_i}(p),l=1,\ldots,k.$$

Then

$$dh_p(u)=\left[ \begin{array}{c} \displaystyle\sum_{i=1}^{k+1}u_i\cfrac{\partial h_1}{\partial x_i}(p)\\ \vdots \\ \displaystyle\sum_{i=1}^{k+1}u_i\cfrac{\partial h_{k+1}}{\partial x_i}(p)\\ \end{array} \right].$$

As $c'(t)=(c_1'(t),\ldots,c_{k+1}'(t))$, we have $$\left<{dh_p(u),c'(t)}\right>=\Phi_{k+1}(t,u)=\displaystyle\sum_{i,j=1}^{k+1}\cfrac{\partial h_j}{\partial x_i}(p)u_i\cdot c_j'(t).$$

hence

$$\begin{array}{l l l} \Phi_l(t,u)&=&\displaystyle\sum_{i=1}^{k+1}u_i\cfrac{\partial F_l}{\partial x_i}(h(p)),\qquad l=1,\ldots,k.\\ & & \\ \Phi_{k+1}(t,u)&=&\displaystyle\sum_{i,j=1}^{k+1}u_i\cfrac{\partial h_j}{\partial x_i}(p)\cdot c_j'(t) \end{array}.$$

Then

$$\cfrac{\partial \Phi_l}{\partial u_j}(t,u)=\cfrac{\partial F_l}{\partial x_j}(h(p)),\qquad \mbox{ for }l=1,\ldots,k,j=1,\ldots,k+1,$$

hence

$$\left[\cfrac{\partial\Phi_l}{\partial u_j}(t,u)\right]_{1\leq j\leq k+1,\\ 1\leq l\leq k}=\left[\cfrac{\partial F_l}{\partial x_j}(h(p))\right]_{1\leq j\leq k+1,\\ 1\leq l\leq k}=dF_{c(t)}.$$

And also for $i=1,\ldots,k+1$ $$\cfrac{\partial\Phi_{k+1}}{\partial u_i}(t,u)=\displaystyle\sum_{j=1}^{k+1}\cfrac{\partial h_j}{\partial x_i}(p)\cdot c_j'(t)=\left(\cfrac{\partial h_1}{\partial x_i}(p),\ldots,\cfrac{\partial h_{k+1}}{\partial x_i}(p)\right)\cdot c'(t).$$

As $$dh_p=\left[ \begin{array}{ccc} \cfrac{\partial h_1}{\partial x_1}(p) &\cdots& \cfrac{\partial h_1}{\partial x_{k+1}}(p) \\ \vdots & \dots & \vdots \\ \cfrac{\partial h_{k+1}}{\partial x_1}(p) &\cdots& \cfrac{\partial h_{k+1}}{\partial x_{k+1}}(p) \end{array} \right],$$

then

$$(dh_p)^t=\left[ \begin{array}{ccc} \cfrac{\partial h_1}{\partial x_1}(p) &\cdots& \cfrac{\partial h_{k+1}}{\partial x_{1}}(p) \\ \vdots & \dots & \vdots \\ \cfrac{\partial h_{1}}{\partial x_{k+1}}(p) &\cdots& \cfrac{\partial h_{k+1}}{\partial x_{k+1}}(p) \end{array} \right],$$

so

$$\begin{array}{l l l} (dh_p)^t(c'(t))&=&\left[ \begin{array}{ccc} \cfrac{\partial h_1}{\partial x_1}(p) &\cdots& \cfrac{\partial h_{k+1}}{\partial x_{1}}(p) \\ \vdots & \dots & \vdots \\ \cfrac{\partial h_{1}}{\partial x_{k+1}}(p) &\cdots& \cfrac{\partial h_{k+1}}{\partial x_{k+1}}(p) \end{array} \right]\left[ \begin{array}{c} c'_1(t)\\ \vdots \\ \vdots \\ . \\ c'_{k+1}(t) \end{array} \right]\\ & &\\ &=&\left[ \begin{array}{c} \cfrac{\partial\Phi_{k+1}}{\partial u_1}(t,u)\\ \vdots \\ \cfrac{\partial\Phi_{k+1}}{\partial u_{k+1}}(t,u) \end{array} \right] \end{array}.$$

Hence, we get that

$$z=\left(\cfrac{\partial\Phi_{k+1}}{\partial u_1}(t,u),\dots, \cfrac{\partial\Phi_{k+1}}{\partial u_{k+1}}(t,u)\right),$$

$z\in Im((dh_p)^t)$ and, if $(u,t)\in\Phi^{-1}(w,0)$ i do not how to use what i have gotten.

• Um... I guess that "Suponha" word is suppose, but something tells me your spanish or portuguese happened in there. – Patrick Da Silva May 26 '14 at 3:23
• Thank you, I have edited it – Miguemate May 26 '14 at 3:38
• I have writen my advance – Miguemate May 28 '14 at 0:56