# Proof the degree of a reflection through a hyperplane is −1.

This seems intuitively true, but it seems impossible to work it out according to the definition on Guillemin and Pollack’s Differential Topology Page 108, tracing back t0 107 and 100.

Page 108:

We define the degree of an arbitrary smooth map $f: X \to Y$ to be the intersection number of $f$ with any point $y$, $\deg(f) = I(f,\{y\})$.

Page 107

If $f: X \to Y$ is transversal to $Z$, then $f^{-1}(Z)$ is finite number of points, each with an orientation number $\pm 1$ provided by the preimage orientation. Define the intersection number $I(f,Z)$ to be the sum of these orientation numbers.

Page 100:

Let $f: X \to Y$ be a smooth map with $f \pitchfork Z$ and $\partial f \pitchfork Z$, where $X,Y,Z$ are oriented and the last two are boundaryless. We define a preimage orientation on the manifold-with-boundary $S = f^{-1}(Z).$

Then the preimage orientation will be determined by
$$df_xN_x(S;X) \oplus T_z(Z) = T_z(Y),$$
where $N_x(S; X)$ be the orthogonal complement to $T_x(S)$ in $T_x(X)$.

I looked into the basic stuff, but eventually the definitions constitute the claim does not go through! Could anyone give me some help to clear this out? Thanks.

#### Solutions Collecting From Web of "Proof the degree of a reflection through a hyperplane is −1."

Does this make sense (I can’t guarantee I didn’t make a mistake in the orientation of a point business):

1) I assume by “reflection across a hyperplane”, you mean a map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$. First, there is no difference if we assume the hyperplane is $\{x_{1}=0\}$.

2) According to the definition on page 108, pick $y$ as the origin (doesn’t really matter which one as all values are regular).

3) According to the definition on page 107, $\deg(f)=I(\mathbb{R}^{n},\{y\})$, which is the sum of the intersection numbers of the points in $f^{-1}(y)$, which is just one point.

4) As for the orientation business, an equivalent definition for whether you get negative 1 or positive 1 is whether $df$ has determinant positive or negative at $f^{-1}(y)$.

If you want to use their definition, we have to think about what an orientation of a point means, and quite honestly, I just tried looking this up in wikipedia (http://en.wikipedia.org/wiki/Orientation_(vector_space)).

From that webpage, it seems like the orientation of a point is a map from ordered basis to $\{\pm 1\}$, so the “usual” orientation of a point would send the basis (empty set) of the tangent space to 1.

From the definition of preimage orientation you wrote, if we want to find if the basis $B=\emptyset$ of the tangent space of the point is positively oriented, we pick $B’=N_{f^{-1}(y)}(\{y\},\mathbb{R}^{n})$, which will just be a negatively oriented basis for the tangent space of $\mathbb{R}^{n}$.

So our $B’$ satisfies the condition that $df_{f^{-1}(y)}(B’)=T_{y}\mathbb{R}^{n}$ and the image is positively oriented.

Then we ask ourselves if $B\cup B’$ is positively or negative oriented in $\mathbb{R}^{n}$ (and it is negatively oriented).