Bijection between fundamental group and path connected components

Question

Please how can I show that $\pi_0 (X,x) = [( \mathbb{S}^0, 1) ,(X, x)]$ is in bijective correspondence with the path connected components of X.

Answer 1

$S^0=\{-1,+1\}$. Hence the elements of $\pi_0(X,x)$ are homotopy classes $[f]$ of continuous maps $f:\{-1,+1\}\rightarrow X$ such that $f(1)=x$. Define the following function: $$ G:\pi_0(X,x)\rightarrow CC(X)::[f]\mapsto X_{f(-1)}, $$ where $CC(X)$ is the set of the connected components of $X$, and $X_{f(-1)}$ denotes the connected component of the point $f(-1)\in X$, i.e. the union of all connected subsets of $X$ containing $f(-1)$.

Notice that $G$ is well defined, indeed if you have $[f]=[g]$, $f(-1)$ and $g(-1)$ must be in the same connected component (otherwise it would be impossible to continuously deform $f$ into $g$).

Moreover, $G$ is injective, indeed, if $X_{f(-1)}=X_{g(-1)}$ then $f$ and $g$ can be continuously deformed into each other since $f(1)=g(1)=x$, which implies $[f]=[g]\in\pi_0(X,x)$.

Finally, $G$ is surjective, indeed, given a connected component $C\subseteq X$, the function $f:\{-1+1\}\rightarrow X$ defined by $f(1)=x$, $f(-1)=c$, with $c\in C$ is mapped to $C$ by $G$.