Quotient space of $\{\theta\in\mathcal D(\mathbb R^d,\mathbb R^d):\langle\theta(x),\nu_{∂Ω}(x)\rangle=0\text{ for all }x\in ∂Ω\}$

Question

The following claim can be found in Introduction to Shape Optimization (p. 58), but is missing rigor:

Let $d\in\mathbb N$, $\Omega$ be a $d$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary, $\nu_{\partial\Omega}$ denote the outward-pointing unit normal field on $\partial\Omega$ and $$\mathcal V(\Omega):=\{\theta\in\mathcal D(\mathbb R^d,\mathbb R^d):\langle\theta(x),\nu_{\partial\Omega}(x)\rangle=0\text{ for all }x\in\partial\Omega\}.$$

The claim is that $$\mathcal D(\mathbb R^d,\mathbb R^d)/\mathcal V(\Omega)\to C_c^\infty(\partial\Omega)\;,\;\;\;[\theta]\mapsto\langle\left.\theta\right|_{\partial\Omega},\nu_{\partial\Omega}\rangle\tag1$$ is an isomorphism. How can we show that?

It's been a while, since I thought about quotient spaces. What would be "canonical" surjection for this quotient space?

Answer 1

I think the question should be reformulated in order to actually make sense:

Therefore, let $k\in\{1,\ldots,d\}$ and $\alpha\in\mathbb N\uplus\{\infty\}$ and assume that, more generally, $\Omega$ is a $k$-dimensional embedded $C^\alpha$-submanifold of $\mathbb R^d$ with boundary.

We know that

1. if $\partial\Omega$ is $\mathbb R^d$-closed, then every $C^{\alpha-1}$-differentiable map $\partial\Omega\to F$, where $F$ is a $\mathbb R$-Banach space, has an extension to a $C^{\alpha-1}$-differentiable map $\mathbb R^d\to F$;
2. if $\partial\Omega$ is $\mathbb R^d$-compact, such an extension can be chosen to be compactly supported.

So, if I'm not missing anything, we need to assume that $\partial\Omega$ is $\mathbb R^d$-compact. Now let $E$ be an arbitrary map from $C^{\alpha-1}(\partial\Omega)$ to $C_c^{\alpha-1}(\mathbb R^d)$ with $$\left.E(f)\right|_{\partial\Omega}=f\;\;\;\text{for all }f\in C^{\alpha-1}\tag2$$ and $\overline\nu_{\partial\Omega}\in C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)$ be an extension of $\nu_{\partial\Omega}\in C^{\alpha-1}(\partial\Omega,\mathbb R^d)$. Let $$P:C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)\to C_c^{\alpha-1}(\partial\Omega)\;,\;\;\;\theta\mapsto\left\langle\left.\theta\right|_{\partial\Omega},\nu_{\partial\Omega}\right\rangle.$$ Note that $$\left(P\circ E\overline\nu_{\partial\Omega}\right)(f)=f\;\;\;\text{for all }f\in C^{\alpha-1}(\partial\Omega)\tag3$$ and hence $$C^{\alpha-1}(\partial\Omega)\to C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)\;,\;\;\;f\mapsto E(f)\overline\nu_{\partial\Omega}\tag4$$ is a right inverse of $P$; hence $P$ is surjective. Moreover, $P$ is easily seen to be linear (and continuous) and, by definition, $$\ker P=\left\{\theta\in C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d):\left\langle\left.\theta\right|_{\partial\Omega},\nu_{\partial\Omega}\right\rangle=0\right\}.$$ Thus, by the fundamental theorem on homomorphisms, $$\hat P:C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)/\ker P\to\operatorname{im}P=C_c^{\alpha-1}(\partial\Omega)\;,\;\;\;\theta+\ker P\mapsto P(\theta)=\left\langle\left.\theta\right|_{\partial\Omega},\nu_{\partial\Omega}\right\rangle$$ is an isomorphism.