Theorem. Let $x_0\in X,$ and let $X_0$ be the path component of $X$ containing $x_0.$ Then $\pi_1(X_0,x_0)\approx\pi_1(X,x_0).$
What is this theorem telling?
Is it telling that the fundamental group of whole space $X$ based at $x_0$ is isomorphic to the fundamental group of $X_0$ based at $x_0$?
How is that it is true? If you take the $\pi_1(S^1,1)$ and $\pi_1(S^n,1)$ and use this theorem for each of them, would not be a contradiction?
Like this $π_1(S^1,1)≈π_1(X,1)$ and $π_1(S^n,1)≈π_1(X,1).$ Hence $π_1(S^1,1)≈π_1(S^n,1).$
(sorry for not to be clear)
If you could give me an intuitively explanation, that would be enough. Thank you.
Let’s use your example of a point $p$ on the sphere $S^n$ and show why this isn’t a contradiction. Your theorem states
$$\pi_1(X_0, x_0) = \pi_1(X,x_0).$$
Here, we have $x_0 = p$ and $X = S^n$. By definition, the subspace $X_0$ is the path-connected component of $X$ containing $x_0$. We don’t get to pick what $X_0$ is — we only get to pick the base point $p$ and then $X_0$ is given to us. Since $S^n$ is a path-connected space, i.e. it has one path-connected component, this implies that $X_0 = S^n$. Thus, your theorem says in this case that
$$\pi_1(S^n, p) = \pi_1(S^n,p).$$
Hopefully this equality is believable!