Theorem. Let $X$ and $Y$ be arbitrary topological spaces with $Y$ (path) connected. If there are maps $f:X\rightarrow Y$ and $g:Y\rightarrow X$ such that $g\circ f:X\rightarrow X$ is homotopic to the identity map, then $X$ is (path) connected.

Proof: Let $f:X\rightarrow Y$ and $g:Y\rightarrow X$ be maps satisfying theorem's assumption. Furthermore let $X=\bigcup X_i$ be a decomposition of $X$ into (path) connected components. Since $Y$ is (path) connected, then $g(Y)\subseteq X_i$ for some $i$ . Thus $(g\circ f)(X)\subseteq X_i$ . Now let $H:I\times X\rightarrow X$ be the homotopy from $g\circ f$ to the identity map. Let $\alpha_{x}:I\rightarrow X$ be a path defined by the formula: $\alpha_{x}(t)=H(t,x)$ . Since for all $x\in X$ we have $\alpha_{x}(0)\in X_i$ and $I$ is path connected, then $\alpha_{x}(I)\subseteq X_i$ . Therefore $H(I\times X)\subseteq X_i$ , but $H(\{1\}\times X)=X$ which implies that $X_i=X$ , so $X$ is (path) connected. $\square$

Straightforward application of this theorem is following:

Corollary. Let $X$ and $Y$ be homotopy equivalent spaces. Then $X$ is (path) connected if and only if $Y$ is (path) connected.