Let $S_ 0$ be the space with $2$ points and the discrete topology. Find [$S_ 0$ , $X$] for an arbitrary space $X$. $[X,Y]=\{f:X\to Y,f$ continuous $\}/\sim$ where $\sim$ is the homotopic equivalence.
I have only proved that if $X$ is contractible then [$S_ 0$ , $X$] is singleton & if $p_1,P_2\in X$ be connected by a path then $\bar C_{p_1}=\bar C_{p_2}$ where $C_{p_1}(x)=p_1 \forall x\in S^0$. So any help will be appreciated.
Let $S^0 = \{1,-1\}$. Then $S^0 \times I = \{1,-1\}\times I = \{1\} \times I \cup \{-1\} \times I$ is basically a disjoint union of two copies of the unit interval $I$. Hence a homotopy $H$ between two maps $f,g \colon S^0 \to X$ "consists" of two paths in $X$, one, $t\mapsto H(1,t)$, connecting $f(1)$ and $g(1)$, the other, $t \mapsto H(-1,t)$, connecting $f(-1)$ and $g(-1)$.
Conversely, if $\alpha\colon I \to X$ is a path connecting $f(1)$ to $g(1)$, and $\beta\colon I \to X$ is a path connecting $f(-1)$ to $g(-1)$, then we get a homotopy between $f$ and $g$ by setting
$$H(1,t) = \alpha(t),\quad H(-1,t) = \beta(t).$$
So $f$ and $g$ are homotopic if and only if $f(1)$ can be connected to $g(1)$ by a path and $f(-1)$ can be connected to $g(-1)$ by a path. In other words, if and only if $f(p)$ and $g(p)$ lie in the same path component of $X$ for every $p\in S^0$.
This gives a bijection between $[S^0,X]$ and the set of ordered pairs of path components of $X$ via that map $f \mapsto (C(f(1)), C(f(-1)))$, where $C(x)$ denotes the path component of $x$ in $X$.