For $n$ even, antipodal map of $S^n$ is homotopic to reflection and has degree $-1$? How do I see that for $n$ even, the antipodal map of $S^n$ is homotopic to the reflection$$r(x_1, \dots, x_{n+1}) = (-x_1, x_2, \dots,  x_{n+1}),$$and therefore has degree $-1$? Thanks for your time.
Progress. I think we want to create a continuous transformation from$$ \begin{pmatrix} -1 & 0 & \ldots & 0 \\ 0 & -1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & -1 \end{pmatrix}$$to$$ \begin{pmatrix} -1 & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 \end{pmatrix}?$$
 A: We start by showing the map $x=(x_1,x_2,\ldots ,x_n)\mapsto (-x_1,-x_2,\ldots, x_n)$ where we flip the sign on the first two coordinates is homotopic to the identity. Suppose $x_1^2+x_2^2=r(x)^2$ for some $0\leq r(x)\leq 1$. Say $(x_1,x_2)=(r(x)\cos \theta(x),r\sin\theta(x))$.Note that $r(x)$ and $\theta(x)$ are continuous functions of $x$, provided you think of $\theta(x)$ as an element of the circle so $\theta=\theta+2n\pi$.
 Consider the homotopy $$(x_1,x_2,\ldots ,x_n)\mapsto (r\cos (\theta+t\pi),r\sin(\theta+t\pi),x_3,\ldots,x_n).$$
This is a rotation of the entire sphere which fixes the last $n-2$ coordinates and reverses the sign of the first two coordinates.
Now you can perform a similar homotopy to reverse the sign of any pair of coordinates, which is all you need to complete the proof.
A: Note that $\rho = \rho_0 \cdots \rho_n$, where $\rho_i(x_0, \dots, x_n) = (x_0, \dots, x_{i-1}, -x_i, x_{i+1}, \dots, x_n)$. Since $\rho_i$ is conjugate to $\rho_j$ via a homeomorphism $g\in SO(n)$, we have $\deg \rho = (\deg \rho_0) \cdots (\deg \rho_n) = (\deg \rho_0)^{n+1}$. But $(\deg \rho_0)^2 = (\deg \rho_0^2) = 1$, so $\deg \rho = \deg \rho_0$ by the fact that $n$ is even.
A: We claim for any sphere $S^n$, there is a homotopy between the identity and the double reflection$$(x_0, x_1, x_2, \dots, x_n) \mapsto (-x_0, -x_1, x_2, \dots, x_n).$$To see this, define$$\ell: S^n \to \mathbb{R},\text{ }r(x_0, \dots, x_n) = \sqrt{x_0^2 + x_1^2},$$and$$\theta:S^n \to [0, 2\pi),\text{ }\theta(x_0, \dots, x_n) = \text{Arg}(x_0 + ix_1).$$Since$$(x_0, x_1) = (\ell(x)\cos(\theta(x)), \ell(x)\sin(\theta(x))),$$we have the homotopy$$H: S^n \times [0, \pi] \to S^n,\text{ }H(x, t) = (\ell(x)\cos(\theta(x) + t),\ell(x)\sin(\theta(x) + t),x_2, \dots, x_n)$$taking$$H(x, 0) = (x_0, x_1, x_2, \dots, x_n) \text{ to }H(x, \pi) = (-x_0, -x_1, x_2, \dots, x_n).$$In fact, there is nothing special about the indices $0$ and $1$, as this argument shows there is a homotopy between the identity and any double reflection. In particular, if $n$ is even, we have a homotopy between the identity and$$(x_0, x_1, \dots, x_n) \mapsto (x_0, -x_1, \dots, -x_n).$$Composing this homotopy with the antipodal map gives the desired homotopy between the antipodal map and the single reflection $r$.
