Let $d\in\mathbb N$, $\alpha\in\mathbb N\uplus\{\infty\}$, $\Omega$ be a bounded $d$-dimensional properly embedded $C^\alpha$-submanifold of $\mathbb R^d$ with boundary, $\nu_{\partial\Omega}$ denote the outward-pointing unit normal field on $\partial\Omega$, $$P:C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)\to C_c^{\alpha-1}(\partial\Omega)\;,\;\;\;\theta\mapsto\left\langle\left.\theta\right|_{\partial\Omega},\nu_{\partial\Omega}\right\rangle$$ and $g\in C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)'$ with $\ker P\subseteq\ker g$.

I would like to conclude that there is a $\gamma\in C_c^{\alpha-1}(\partial\Omega)'$ with $$g=\gamma\circ P\tag1.$$

I was able to show that $P$ is a linear (continuous) surjection and hence, 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.

Now let $$\pi:C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)\to C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)/\ker P\;,\;\;\;\theta\mapsto\theta+\ker P.$$

We should be able to construct $\gamma$ using $\pi$ and $\hat P$. Maybe we need to apply the fundamental theorem on homomorphisms to $g$, but instead of "dividing" by $\ker g$, we should somehow "divide" by the (closed) subspace $\ker P$ of $\ker g$. But I wasn't able to figure out the details.

At the end we should have obtained a linear map $\gamma$. How can we show that this map is continuous as well?


The context here is superfluous, so let us strip the cumbersome notation. Let $K$ be a field, $P\colon V\rightarrow W$ be a surjective linear map between $K$-vector spaces and $g\colon V\rightarrow K$ be a linear map such that $\ker P\subseteq\ker g$. You are asking why there exists a linear map $\gamma\colon W\rightarrow K$ such that $\gamma\circ P=g$.

Let $\pi\colon V\rightarrow V/\ker P$ be the quotient map. As you note, $P$ factors through the quotient as an isomorphism $\tilde{P}\colon V/\ker P\rightarrow W$ (by the isomorphism theorem). Furthermore, $\ker P\subseteq\ker g$, so $g$ factors through the quotient as a linear map $\tilde{g}\colon V/\ker P\rightarrow K$ (by the universal property of the quotient). Then $\gamma:=\tilde{g}\circ\tilde{P}^{-1}\colon W\rightarrow K$ does the job.

  • $\begingroup$ Thank you for your answer. (a) I'm not sure what exactly the "universal property of the quotient" is that you are referring to (maybe you can clarify), but it's easy to see that $$V/A\to K\;,\;\;\;v+A\mapsto g(v)$$ is a linear surjection for all subspaces $A$ of $\ker g$. (b) If I'm not missing anything, the surjectivity of $P$ is not needed. Even without that assumption $\gamma$ is a well-defined map from $\operatorname{im}P$ to $\operatorname{im}g$. (c) The only thing left is the continuity. How can we prove that? $\endgroup$
    – 0xbadf00d
    Oct 12 '20 at 4:38
  • $\begingroup$ (d) BTW, you didn't used the $\pi$ you've declared. I guess the point is that $\tilde P^{-1}\circ P=\pi$. $\endgroup$
    – 0xbadf00d
    Oct 12 '20 at 5:56
  • $\begingroup$ (a) The universal property of the quotient is that every other map whose kernel contains $\ker P$ factors through $\pi\colon V\rightarrow V/\ker P$. The map you mention here is the required one (thought it's not necessarily surjective). (b) You are correct. $\gamma$ is well-defined as a map $\operatorname{im P}\rightarrow K$ and then can be extended to a map $W\rightarrow K$. If $P$ is surjective, however, $\gamma$ is uniquely determined. $\endgroup$
    – Thorgott
    Oct 12 '20 at 19:25
  • $\begingroup$ (c) If the base field is $\mathbb{R}$, all the vector spaces are Banach spaces and the given maps are continuous, then $\gamma$ is continuous as composition of continuous maps. The factor maps $\tilde{g},\tilde{P}$ are continuous by the definition of the quotient topology and $\tilde{P}^{-1}$ is continuous by the Bounded Inverse theorem. (If we return to (b) and you don't assume $P$ to be surjective, the extension I mentioned can be justified with the Hahn-Banach theorem). (d) Indeed, $\pi$ appears when you write out why $\gamma$ works. $\endgroup$
    – Thorgott
    Oct 12 '20 at 19:26

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