My professor mentioned that every finite topological space is homotopy equivalent to its Kolmogorov quotient, but it seems to me that this should be true for any topological space.
Let $X$ be a topological space and let $Y$ be its Kolmogorov quotient (that is, $Y=X/\sim$ under the quotient topology where $x_1\sim x_2$ iff for all open sets, $U$, $x_1\in U$ if and only if $x_2\in U$). Let $f:X\to Y$ be the canonical surjection and let $g:Y\to X$ be any choice function. Then $f$ and $g$ are continuous and $f\circ g=id_Y$. Define $H:X\times[0,1]\to X$ by $H(x,0)=(g\circ f)(x)$ and $H(x,t)=x$ for any $t>0$. Since $g(f(x))$ is topologically indistinguishable from $x$, $(g\circ f)^{-1}(U)=U$ for any open set $U\subset X$. So, $H^{-1}(U)=U\times[0,1]$ for any open set $U$ which implies $H$ is continuous. Thus $g\circ f\simeq id_X$ so that $X$ and $Y$ are homotopy equivalent. Did I make a mistake somewhere?
Best Answer
Yes, this is correct: the Kolmogorov quotient map of any topological space is a homotopy equivalence, with any section as a homotopy inverse.