In Topology: A Categorical Approach it's stated that the Hausdorff property is topological but not homotopy invariant. The only example that I know of a "naturally occurring" non-Hausdorff space is the line with two origins, and the only thing I can think to do is use the quotient procedure that created it, but I don't think that produces a homotopy equivalence. Is there an elementary example of a homotopy equivalence between a Hausdorff and non-Hausdorff space? Or am I wrong in thinking that the quotient procedure $(x,a) \sim (x,b)$ if $x \neq 0$ does not give a homotopy equivalence?
Hausdorff property is not homotopy invariant
general-topologyhomotopy-theory
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Let $X = \mathbb{R}^n / \sim$, and consider the projection map $\pi : \mathbb{R}^n \to X$ which maps each point to its equivalence class. By definition of the quotient topology, $\pi$ is continuous. If we let $E = S^{n-1} \cup \{0\}$ consist of the unit sphere in $\mathbb{R}^n$ together with the origin, then $E$ is compact in $\mathbb{R}^n$, and $\pi(E) = X$ (as $E$ contains at least one point from every equivalence class). Hence $X$ is the continuous image of a compact set.
To see it is not Hausdorff, notice that the only open set in $X$ which contains the equivalence class $[0]$ is $X$ itself. Indeed, if $U$ is open in $X$ and contains $[0]$, then $\pi^{-1}(U)$ contains a neighborhood of $0$ in $\mathbb{R}^n$; in particular it contains a point of every equivalence class. So in fact $U$ contains every equivalence class in $X$. Now this prevents $X$ from being Hausdorff, since we cannot find disjoint open neighborhoods of $[0]$ and any other point.
$\def\RR{\mathbb{R}}$There is no separation condition which will do the job. That's a vague statement, so here is a precise one: There is a subset of $\mathbb{R}^2$ which (equipped with the subspace topology) does not have condition $\dagger$.
Proof: Let $A$ and $B$ be two disjoint dense subsets of $\mathbb{R}$, neither of which contains $0$. (For example, $\mathbb{Q} +\sqrt{2}$ and $\mathbb{Q}+\sqrt{3}$.) Let $$X = (A \times \RR_{\geq 0}) \cup (B \times \RR_{\leq 0}) \cup (\{0\} \times \RR_{\neq 0}) \subset \RR^2.$$ Define $(x_1,y_1)$ and $(x_2, y_2)$ to be equivalent if $x_1=x_2$ and, in the case that $x_1=x_2=0$, that $y_1$ and $y_2$ have the same sign.
Verification that this is a closed equivalence relation: $X^2$ is a metric space, so we can check closure on sequence. Let suppose we have a sequence $(x_n, y_n) \sim (x'_n, y'_n)$ with $\lim_{n \to \infty} x_n=x$, $\lim_{n \to \infty} y_n=y$, $\lim_{n \to \infty} x'_n=x'$ and $\lim_{n \to \infty} y'_n=y'$. We must verify that $(x,y) \sim (x',y')$. First of all, we have $x_n = x'_n$, so $x=x'$ and, if $x=x' \neq 0$, we are done. If $x=x'=0$, we must verify that $y$ and $y'$ have the same sign. But $y_n$ and $y'_n$ weakly have the same sign for all $n$, so they can't approach limits with different signs.
Verification that $X/{\sim}$ is not Hausdorff: We claim that no pairs of open sets in $X/{\sim}$ separates the images of $(0,1)$ and $(0,-1)$. Suppose such open sets exist, and let $U$ and $V$ be their preimages in $X$. Then there is some $\delta$ such that $(A \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset U$ and $(B \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset B$. Then $U \cap \RR \times \{ 0 \}$ is an open set which contains $(-\delta, \delta)$. By the density of $B$, there must be a point of $B \cap (- \delta, \delta)$ in $U \cap \RR$, and then this gives an intersection between $U$ and $V$.
Best Answer
Here's an example of a non-Hausdorff space:
Let $X$ be your favourite nontrivial Hausdorff space. Let $Y=X\times\left\{1,2\right\}$ be the product space of $X$ with the chaotic topological space $\left\{1,2\right\}$, where the only open subsets of $\left\{1,2\right\}$ are $\varnothing$ and $\left\{1,2\right\}$ itself.
More informally, $Y$ consists of two copies of $X$, namely $X_1=X\times\left\{1\right\}$ and $X_2=X\times\left\{2\right\}$. Obviously, $Y$ is not Hausdorff.
But $X$ and $Y$ are homotopy invariant: Take the maps $f\colon X\to Y$ which maps $X$ to the first copy of it inside $Y$: $f(x)=(x,1)$; and let $g\colon Y\to X$ maps whatever element of any copy of $X$ to the same element of $X$: $g(x,i)=x$.
Then $gf=\operatorname{id}_X$, and the function $$H\colon Y\times[0,1]\to Y,\qquad H((x,i),t)=\begin{cases}(x,1)&\text{ if }t=0\\(x,i)&\text{ if }t>0\end{cases}$$ is a homotopy from $fg$ to $\operatorname{id}_Y$. [The only nontrivial part is continuity of $H$: Note that $H^{-1}(U)=U\times[0,1]$ for all $U\subseteq Y$ open.]