Is the quotient topological space $[0,1]/(0,1)$ contractible?

Question

The question is already in the title. Is the quotient space $[0,1]/(0,1)$ contractible ?

Here I have three points, but not in the discrete topology. Further, the only open sets are $\{[(0,1)]\}$ and the whole space. I don't know if that helps.

I feel like this should be easy, but I can't figure it out.

Edit:
I followed @Marcos hint.
But I find a counterexample where $(0,1) \to [0,1]$ do not have the homotopy extension property. If I take $Y:= \{0\} \times [0,1] \cup (0,1) \times [0,1]$ with the subspace-topology of $\mathbb{R}^2$, $H:(0,1) \times [0,1] \to Y$ as the identity, and $\tilde{H}_0(t) = (t,0)$ this cannot be extended to a homotopy $\tilde{H}:[0,1] \times [0,1] \to Y$.

Answer 1

The quotient space $X = [0, 1] / (0, 1)$ consists of three points that we will denote $0, 1$ and $a$, where $a$ is the image of any $x \in (0, 1)$ under the quotient map $[0, 1] \to [0, 1] / (0, 1)$. The topology of $X$ is not Hausdorff, its closed sets are $\emptyset, X, \{0\}, \{1\}$ and $\{0, 1\}$.

One can prove that the map $h \colon X \times [0, 1] \to X$ defined by

$$h(x, t) = \begin{cases} x & \text{if } t = 0 \\ a & \text{if } t > 0 \end{cases}$$

is a deformation retraction of $X$ onto $\{a\}$. We need to prove the continuity of $h$. For this, we only need to prove that the inverse image of the closed sets $\{0\}$ and $\{1\}$ under $h$ are closed. Indeed, $h^{-1}(\{0\}) = \{0\} \times \{0\}$ and $h^{-1}(\{1\}) = \{1\} \times \{0\}$.

Answer 2

You can get intuition for the other answer as follows. Suppose X is a topological space and A a subspace. Suppose one has a deformation retraction of X onto a subspace of A, so that for all $t$ we have $h_t(A) \subset A$. Then quotienting by $A$ we have a map $h: [0,1] \times X \to X/A$ which is the projection for $t = 0$ and a constant map for $t = 1$. Further, if one defines $\sim$ to be the relation $(t,a) \sim (t,a')$ whenever $a, a' \in A$ (and no other points are equivalent to anything other than themselves), this map is constant on $\sim$-equivalence classes, and descends to a map $\big([0,1] \times X\big)/\sim \to X/A$.

When you are lucky, the quotient by the relation above is homeomorphic to $[0,1] \times X/A$ and you have defined a deformation retraction of X/A to a point. This is the case eg when X is compact and A is closed.

Here this is probably not the case but it does suggest the right approach. Let me work with $X = [-1, 1]$ and $A = (-1,1)$ for ease of notation. Then $h_t(s) = (1-t)s$ is an example of a mapping as above. When you quotient by A everywhere this map descends to precisely the map defined in the other answer, and you can check it's continuous directly instead of worrying about whether the quotient topology is correct.