Notes to my Cat---Path Connected implies Connected, an elegant proof.

Notation. $2$$2$ is the set $\{0,1\}$$\{0,1\}$ with discrete topology

Lemma. Let $X$$X$ be a topological space, then $X$$X$ is connected iff ${\mathrm{Hom}}_{\mathrm{Top}}(X,2)=\{f,g\}$$\mathrm{Hom}_{\mathrm{Top}}(X,2)=\{f,g\}$, or there is no continuous surjection from $X$$X$ to $2$$2$.

Proof. We only need to prove that $X$$X$ is disconncted iff there exists a continuous surjection from $X$$X$ to $2$$2$.

Let $f:X\to 2$$f:X\to 2$ be a continuous surjection, then $f^{-1}(0),f^{-1}(1)$$f^{-1}(0),f^{-1}(1)$ is two nonempty disjoin open set and $f^{-1}(0)\cup f^{-1}(1)=X$$f^{-1}(0)\cup f^{-1}(1)=X$.

Conversely let $X$$X$ be disconnected, $X=U\cup V,U\cap V=\mathrm{\varnothing }$$X=U\cup V,U\cap V=\empty$ for two no empty open sets $U,V$$U,V$. Then $f(U)=0,f(V)=1$$f(U)=0,f(V)=1$ is a continuous surjection. $◻$$\square$

Proposition. $X$$X$ is path connected implies $X$$X$ is connected.

Proof. Suppose $X$$X$ is path connected but not connected, then there exists a continuous surjection $f:X\to 2$$f:X\to 2$.

Hence there exists a continuous surjection $p:[0,1]\to \{0,1\}$$p:[0,1]\to\{0,1\}$, i.e. $[0,1]$$[0,1]$ is not connected, that is a contradiction. $◻$$\square$

Remark. Let $p^{\prime }:[0,1]\to X$$p':[0,1]\to X$ be the path with $p^{\prime }(0)\in f^{-1}(0),p^{\prime }(1)\in f^{-1}(1)$$p'(0)\in f^{-1}(0),p'(1)\in f^{-1}(1)$. Then $p=f\circ p^{\prime }$$p=f\circ p'$ is a continous surjection from $[0,1]\to 2$$[0,1]\to 2$.