(path) connectness as a homotopy invariant

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

Proof: Let $f:X\to Y$ and $g:Y\to X$ 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 $H:I\times X\to X$ be the homotopy from $g\circ f$ to the identity map. Let ${\alpha }_x:I\to X$ 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. $\mathrm{\square }$

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.