I tried to compute the degree of the reflection map. I'm wondering if I'm right ?
Let the reflection map, $$ r_i: S^{n} \longrightarrow S^{n}, (x_1, ..., x_i,...,x_{n+1}) \longrightarrow (x_1, ..., -x_i, ..., x_{n+1}) $$ Compute the degree of that mapping.
Well, because $S^{n}$ is compact and connected and $r_i$ is a mapping between two manifold of the same dimension, the degree of that mapping is the same as the local degree of that mapping evaluated at any regular value $y \in S^{n}$ by the theorem stating the independance of the local degree with respect to the regular value,
$$ \deg{(r_i)} = \deg{(r_i ; y)} $$
So the first step is to choose a regular value of $S^{n}$.
Because the reflexion $r_i$ is across any plane of equation $x_i = 0$, we can choose in particular a reflexion that is NOT across the plane $\pi \equiv x_1 = 0$.
The point $x^{\ast} = (1, 0, 0, ...)$ is therefore a fixed point of that reflexion : $r_i(x^{\ast}) = x^{\ast}$ where $i \neq 0$.
Then, because $x^{\ast}$ is a fixed point, $r_i$ is a one-to-one mapping in $x^{\ast}$ and then the differential,
$$ d(r_i)_{x^{\ast}}: T_{x^{\ast}}S^{n} \rightarrow T_{r(x^{\ast})}S^{n} = T_{x^{\ast}}S^{n} $$
is an isomorphism. In particular, $d(r_i)_{x^{\ast}}$ is a mapping onto then $x^{\ast}$ is a regular point.
We compute the degree in the corresponding regular value (which is the same as that regular point),
$$ \deg{(r_i ; r_i(x^{\ast}))} = \text{sign}(\det(d(r_i)_{x^{\ast}})) $$
We see that,
$$ d(r_i)_{x^{\ast}} = \begin{pmatrix} 1 & ... & ... & ... & ... &... & 0 \\ ... & ... & ... & ... & ... &... & 0 \\ ... & ... & ... & -1 & ... & ... & 0 \\ ... & ... & ... & ... & ... &... & 0 \\ 0 & ... & ... & ... & ... &... & 1 \end{pmatrix} $$
And,
$$ \text{det}(d(r_i)_{x^{\ast}}) = \prod_{i=1}^{n+1} a_{ii} = -1 $$
Therefore,
$$ \deg{(r_i)} = \deg{(r_i ; r_i(x^{\ast}))} = \text{sign}(-1) = -1 $$
The degree of the reflection map is $-1$.
Is it correct ?