I want to prove that total disconnectedness is invariant under the same homotopy type, i.e. if two spaces $X$ and $Y$ are homotopy equivalent, then $X$ is totally disconnected iff $Y$ is totally disconnected.
My attempt:
Let $f:X\rightarrow Y$ and $g:Y\rightarrow X$ be continuous functions s.t. $f\circ g\simeq id_Y$ and $g\circ f\simeq id_X$.
Suppose $X$ is totally disconnected, and let $Q$ be a component of $Y$. Then $g(Q)$ is constant and $(f\circ g)(Q)\in Q$.
How to prove that $Q$ consist of only one element?
Best Answer
Aren't (a) the interval $X = [-1, 1]$ and (b) the singleton $Y = \{0\}$, both with the inherited topology from $\Bbb R$, homotopy equivalent, with the map $$ f : X \to Y : x \mapsto 0 $$ and $$ g: Y \to X : 0 \mapsto 0 $$ being the equivalences? After all, $f \circ g$ is actually the identity on $Y$, and $g \circ f$ is homotopic to the identity on $X$ via $$ H(x, t) = (1-t)x, $$ where $H(x, 0) = 0$ is the map $f \circ g$, and $H(x, 1) = 1$ is the identity on $X.$
If so, your conjecture has problems, as the space $X$ is not totally disconnected. .