# How to show $(d\pi^{-1})_{\pi(y)}\circ (d\pi)_x:T_xS^n\longrightarrow T_y S^n$ reverses orientation for $n$ even?

Let $\mathbb R\mathbb P^n$ be the quotient manifold $S^{n}/R$ where $R$ is the equivalence relation given by: $$xRy\Leftrightarrow y=x\ \textrm{or}\ y=-x.$$ We know the canonical map $\pi:S^n\longrightarrow \mathbb R\mathbb P^n$ is a local diffeomorphism. How can I show that if $n$ is even then $$(d\pi^{-1})_{\pi(y)}\circ (d\pi)_x:T_xS^n\longrightarrow T_y S^n,$$ reverses orientation whenever $\pi(x)=\pi(y)$ with $x\neq y$?

Your question is equivalent to the statement that the antipodal map of $S^n$ reverses orientation if $n$ is even. All you have to do is prove this for one antipodal pair or points $x,y$, because it will then follow for all antipodal pairs by continuity.
So, using coordinate notation in $\mathbb{R}^{n+1}$, let $x = e_1 = (1,0,0,0,…,0)$ and let $y=-e_1$ in $S^n$. The tangent space to $S^n$ at $x=e_1$ has positively oriented basis $e_2,e_3,…,e_{n+1}$. If you take this $n$-tuple of vectors and parallel transport it around the semicircle where $S^n$ intersects the $e_1,e_2$-plane, you will get a positively oriented basis $-e_2,e_3,…,e_{n+1}$ at $y$. On the other hand, the antipodal map takes the basis $e_2,e_3,…,e_{n+1}$ at $x$ to the basis $-e_2,-e_3,…,-e_{n+1}$ at $y$. The change of basis matrix between these two bases at $y$ is a diagonal matrix with one entry of $1$ and with $n-1$ entries of $-1$. The determinant of that matrix is odd if and only if $n$ is even, so the antipodal map reverses orientation if and only if $n$ is even.