Let $X$ be a topological space. Let $C$ be a path component of $X$ containing the point $x_0 \in X$. Can we say that $\pi_1 (C,x_0) \cong \pi_1 (X,x_0)$?
What I think is as follows $:$
If we can show that all the loops based at $x_0$ are in $C$ then we are through. Suppose $\exists$ a loop at $x_0$ which is not entirely lying in $C$. Then $\exists$ a point $x$ on that loop such that $x \notin C$. But since $x$ is arcwise connected to $x_0$ via that loop and any point in $C$ is connected to $x_0$ so any point in $C$ is connected to $x$. Then $C \cup \{x \}$ becomes a path connected space in $X$ strictly containing $C$, which contradicts the fact that any path component of $X$ is the maximal path connected space in $X$. This proves our claim and we are done with the proof.
Please verify it. Thank you very much.