# Non-closed deformation retract

I am searching for a $T_1$ space $X$ which deformation retracts onto a non-closed subspace $A$. Such a space cannot be Hausdorff as any retract in a Hausdorff space is closed.

I tried some spaces, for example the unit interval with two origins (take $[0,1]×\{0,1\}$ and identify $(x,0)\sim(x,1)$ for $x>0$) and the subspace $A=X-\{(0,1)\}$. Although this is a retract - map $(0,1)$ to $(0,0)$ - it is not a deformation retract.
Note that if $H:X×I\to X$ is the homotopy with $H(x,0)=x$, $H(x,1)=r(x)$ and $H(a,t)=a$ for all $a\in A$, and if $x\in\overline A$, then $H(x,t)$ must be a point in $\overline A$ for every $0<t<1$, and if $H(x,t)\in A$, then $H(x,t)$ and $x$ cannot have disjoint neighborhoods.

Do you have any idea?

You can just refine your counterexample, for example take $[0,1]\times [0,1]$ and identify $(x,t)\sim (x,u)$ if $x>1$ (for any $t,u$).
The quotient space $Y$ that you obtain deformation retracts onto the projection of $[0,1]\times\{0\}$ to the quotient, which is not closed.
Note that you can use the obvious deformation retraction $[0,1]\times [0,1]\to [0,1]\times\{0\}$ and the fact that a deformation retraction on a topological space induces a deformation retraction on a quotient (this is well-known but non-trivial), which in this case is easy.

• Since the deformation retraction respects the relation, it induces a deformation retraction on the quotient. And $Y$ is $T_1$ since the classes are closed. Thanks for your example! – Stefan Hamcke Dec 19 '13 at 21:03
• Exactly. What I wanted to highlight is that if you have a space $X$ and a quotient space $Y=X/\sim$ then a deformation retraction (respecting the relations) induces a map on $X\times [0,1]/\sim'$ (where $(a,t)\sim'(b,t)$ iff $a\sim b$), but the fact that $X\times [0,1]/\sim'$ is canonically homeomorphic to $Y\times [0,1]$ is not trivial. – Mizar Dec 20 '13 at 8:24
• Yah, I am aware of that. It is because for a quotient map $q:X\to Y$ the product $q\times 1_Z$ is a quotient map if $Z$ is locally compact. – Stefan Hamcke Dec 20 '13 at 13:54

Here is another example. Let $X=I=[0,1]$ with the cofinite topology. Then $A=(0,1]$ is an open subspace which is not closed. $X$ is a compact $T_1$ space which retracts onto $A$ by sending $0$ to $1$ in $A$. If we define $$H:X×I\to X\\(x,t)\mapsto\begin{cases} t,\ \ \text{ if }x=0\\x,\ \,\text{ if }x>0 \end{cases}$$ Since $H|_{A×I}$ is continuous and $A$ is open, $H$ is continuous on every point of $A×I$. To check continuity for a point $(0,t)$, let $X-\{x_1,...,x_n\}$ be an open neighborhood of $t$. By $d$ we want to denote the smallest distance between $t$ and any of the points $x_i$. Then $(X-\{x_1,...,x_n\}\cup\{0\})×(t-d,t+d)$ is a neighborhood of $(0,t)$ whose image is contained in $X-\{x_1,...,x_n\}$.