Let $r :\mathbb{C}^{n+1} \rightarrow \mathbb{C}^{n+1} $ be the map $r(z_0, z_1,\ldots, z_n)=(-z_0, z_1,\ldots, z_n)$. $r$ induces a map $\bar r : \mathbb{CP}^n \rightarrow \mathbb{CP}^n $. What is the degree of $\bar r $?
Note: I have shown that there exists no map of $\mathbb{CP}^n \rightarrow \mathbb{CP}^n $ of degree -1 if n is even, using cohomology ring. Since $r$ is an isomorphism of $\mathbb{C}^n$, $\bar r$ is a homeomorphism, so must have degree $\pm 1$. So for even $n$, it is degree 1. but how do I do the case for $n$ odd? My first guess would be $-1$ regardless of dimension since $r$ is a reflection.
Note: I know there are multiple ways to think of complex projective space (even dimensional cells, quotient of $\mathbb{S}^n$, disk with boundary identifications), which one is the best one to use here?
Holomorphic functions are always orientation-preserving, so this is not the reflection you're used to from real linear algebra. The map has degree $1$. If you want to think about it in real coordinates (e.g., looking at the induced map on $S^{2n+1}$), you'll see that it is a composition of two real reflections.