Why does the antipodal map on $S^2$ have degree $-1$?

I’m reading the post here by Arthur to explain that there is no smooth vector field on $S^2$. I don’t understand it very well:

The simplest I can remember off the top of my head is this:

Assume there is such a vector field. Let $v_x$ denote the vector at the point $x$. Now, define the homotopy $H : S^2\times[0,1]\to S^2$ by the following: $H(x,t)$ is the point $t\pi$ radians away from $x$ along the great circle defined by $v_x$. This gives a homotopy between the identity and the antipodal map on $S^2$, which is impossible, since the antipodal map has degree $−1$. Hence there can be no such vector field.

I wonder how can we compute the degree of map defined here? As I imagine, there is probably a theorem saying that all antipodal mapping on $S^2$ has degree $-1$. I wonder why is that? No answers found online.

Solutions Collecting From Web of "Why does the antipodal map on $S^2$ have degree $-1$?"

Consider $S^n$ as the unit sphere centred at zero in $\mathbb{R}^{n+1}$. We can use Proposition $15.21$ of Lee’s Introduction to Smooth Manifolds (second edition), paraphrased below, to construct an orientation form on $S^n$.

Proposition $15.21$: Suppose $M$ is an oriented smooth $n$-manifold, $S$ an immersed hypersurface in $M$, and $N$ is a vector field along $S$ that is nowhere tangent to $S$. Then $S$ has a unique orientation such that for each $p \in S$, $(E_1, \dots, E_{n-1})$ is an oriented basis for $T_pS$ if and only if $(N_p, E_1, \dots, E_{n-1})$ is an oriented basis for $T_pM$. If $\omega$ is an orientation form for $M$, then $\alpha = (i_N\omega)|_S$ is an orientation form for $S$ with respect to this orientation.

In our case, $M = \mathbb{R}^{n+1}$, $S = S^n$, $\omega = dx^1\wedge\dots\wedge dx^{n+1}$ and $N$ is the vector field given by

$$N(x) = \sum_{i=1}^{n+1}x_i\frac{\partial}{\partial x^i}.$$

Using formula $(14.12)$ in Lee, we have

\alpha = (i_N\omega)|_{S^n} &= \left.\sum_{j=1}^{n+1}(-1)^{j-1}dx^j(N)dx^1\wedge\dots\wedge dx^{j-1}\wedge dx^{j+1}\wedge\dots\wedge dx^{n+1}\right|_{S^n}\\
&= \left.\sum_{j=1}^{n+1}(-1)^{j-1}dx^j\left(\sum_{i=0}^{n+1}x_i\frac{\partial}{\partial x^i}\right)dx^1\wedge\dots\wedge dx^{j-1}\wedge dx^{j+1}\wedge\dots\wedge dx^{n+1}\right|_{S^n}\\
&= \left.\sum_{j=1}^{n+1}(-1)^{j-1}x^jdx^1\wedge\dots\wedge dx^{j-1}\wedge dx^{j+1}\wedge\dots\wedge dx^{n+1}\right|_{S^n}.

Now that we have $\alpha$, we can prove the following result.

Proposition: The antipodal map on $S^n$ has degree $(-1)^{n+1}$.

Proof: Let $f$ be the antipodal map on $S^n$. Then we have

f^*\alpha &= \left.\sum_{j=1}^{n+1}(-1)^{j-1}(-x^j)d(-x^1)\wedge\dots\wedge d(-x^{j-1})\wedge d(-x^{j+1})\wedge\dots\wedge d(-x^{n+1})\right|_{S^n}\\
&= \left.\sum_{j=1}^{n+1}(-1)^{j-1}(-x^j)(-dx^1)\wedge\dots\wedge (-dx^{j-1})\wedge (-dx^{j+1})\wedge\dots\wedge (-dx^{n+1})\right|_{S^n}\\
&= (-1)^{n+1}\left.\sum_{j=1}^{n+1}(-1)^{j-1}x^jdx^1\wedge\dots\wedge dx^{j-1}\wedge dx^{j+1}\wedge\dots\wedge dx^{n+1}\right|_{S^n}\\
&= (-1)^{n+1}\alpha.

Therefore, $f$ has degree $(-1)^{n+1}$.

In particular, the antipodal map $S^2 \to S^2$ has degree $(-1)^{2+1} = (-1)^3 = -1$.

As Michael answered, this is a general fact. If $f: S^n \to S^n$ is the antipodal, its degree is $(-1)^{n+1}$. Another way to prove it is to notice that $f$ is the composition of $n+1$ reflections (one for each coordinate). Since the degree of the composition is the product of the degrees and a reflection has degree $-1$, we obtain the result.