Straightening the boundary in concrete examples Let $\Omega \subset \mathbb{R}^d$ be open and with $C^1$ boundary $\Gamma$.
For any given point $x_0 \in \Gamma$ we know there's a neighborhood where
$\Gamma$ is the graph of some $C^1$ function $\gamma : \mathbb{R}^{d - 1}
\longrightarrow \mathbb{R}^d, x' \longmapsto \gamma ( x') = x_d$. We can use
it to straighten the boundary with the local diffeomorphism
$$ T ( x', x_d - \gamma ( x')) = ( x', x_d - \gamma ( x')), $$
and its differential $D T$ has a nice $( d - 1) \times ( d - 1)$ identity
matrix as first block and a bottom row $\nabla T_d = ( - \nabla \gamma, 1)$
which is proportional to the vector $\vec{n}$ normal to $\Gamma$ at each
point, say $c ( x)  \vec{n} ( x) = \nabla T_d ( x)$, where $c ( x) = - \|
\nabla T_d ( x) \|$.
For my calculations in concrete examples with parametrized domains, etc., I
want $\nabla T_d$ to actually be the outward pointing normal: I need this $c (
x)$ to be $- 1$. If I try to impose the condition after constructing $T$, then
I have to integrate expressions which I'm just not capable of. I can try to
throw it at some symbolic integration software, but there has to be some other
way, right? In almost every book on PDEs it's stated that this $T$ may be
normalized so as to have the property I mention. But how?

If $\phi(x')$ denotes the $d$-th component of the normal vector at $(x',\gamma(x'))$, then first of all it is immediate from the graph structure that $\phi(x') \ne 0$. Let $S(x',y_d) = (x',\phi(x')y_d)$, and write $\tilde{T} = S \circ T$. Then $\tilde{T}$ is a $\mathcal{C}^1$ diffeomorphism which straightens the boundary, and it is normalized, as can be checked easily with the chain rule:
$$
DS (x',y_d) = 
\left[
\begin{array}{c|c}
\mathrm{Id} & 0 \\ \hline
\nabla \phi(x') y_d & \phi(x')
\end{array}
\right]
$$
In particular
$$
DS (x',0) = 
\left[
\begin{array}{c|c}
\mathrm{Id} & 0 \\ \hline
0 & \phi (x')
\end{array}
\right]
$$
So on the boundary $\Gamma$ you get
$$
D\tilde{T}(x',\gamma(x')) = DS(x',0) DT(x',\gamma(x'))
=\left[
\begin{array}{c|c}
\mathrm{Id} & 0 \\ \hline
-\phi(x')\nabla \gamma(x') & \phi (x')
\end{array}
\right]
$$
I.e., the last row is a multiple of the outer normal, and since the $d$-th entry is the same, it is equal to the outer normal.
