Consider the equivalence relation on $\mathbb{S}^1$ given by identification of antipodal points $x \sim y \iff x = \pm y$
I have shown that the quotient map $\pi : \mathbb{S}^1 \to \mathbb{R}P^1$ is a $2$-sheeted covering map for projective line, i.e $\pi^{-1}([y]_\sim) = \{y, -y\}$. In a textbook I am reading, the degree of a covering map is defined as the cardinality of a fiber. Here it's two. But I have also read that we can consider the induced homomorphism $$\pi_{*} : \pi_1(\mathbb{S}^1, y) \to \pi_1(\mathbb{R}P^1, [y]) \quad \quad \gamma \mapsto \pi \circ \gamma$$ and since both groups here are $\mathbb{Z}$, we can think of $\pi_{*}$ as multiplication by some integer $n \in \mathbb{Z}$ : $k \mapsto n k$ and that's its degree.
I think both definitions are equivalent, right? How do I show using second definition that $n = 2$? (I know very little group theory)
You have a homeomorphism $\Bbb{R}P^1 \cong S^1$ given by $[z] \mapsto z^2$, so the composition $S^1 \to \Bbb{R}P^1 \cong S^1$ is the map $z \mapsto z^2$. From there it's clear that $\pi_*(1) = 2$.