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Let $X = [0,1]/(0,1)$ and let $\pi: [0,1] \rightarrow X$ be the quotient map. Answer the following questions, proving your assertions:

a) Is $X$ contractible?

We need $s:X \rightarrow \{pt\}$ and $t: \{pt\} \rightarrow X$ with $s \circ t \simeq id_{\{pt\}}$ and $t \circ s \simeq id_{X}$.

$s$ is the obvious map. Define $t$ with $pt \mapsto 0$. We have $s \circ t = id_{\{pt\}}$. Now define $H(x,t) = (t \circ s)(1-t) + id_{X}(t)$, which is a homotopy from $t \circ s$ to $id_{X}$. So Yes, $X$ is contractible.

b) Are $\pi(0)$ and $\pi(1/2)$ dense in $X$?

$\{0\}$ is closed in $X$ because $\pi^{-1}(\{0\})$ is closed in $[0,1]$. So $\overline{\{0\}} \not = X \implies \pi(0)$ is not dense in $X$.

Since $\overline{\pi(1/2)}=\overline{(0,1)} = [0,1]$, $\pi(1/2)$ must be dense in $X$.

c) Consider the complement of $\pi(0)$ in $X$. Is it Hausdorff? Is it connected?

No it's not Hausdorff. For $[1/2]$ and $1$, we need to find $U$ and $V$ open and disjoint in $X$ with $[1/2] \in U$ and $1 \in V$. But we need $V \subseteq X - (0,1)$. So $V = \{1\}$, which is closed (since $\pi(\{1\})$ is not open in $[0,1]$).

It is connected. Suppose we have $X=A \cup B$ for $A,B$ open, nonempty, and disjoint in $X$. Since there are only two elements, the only possibilities for $A$ and $B$ are $(0,1)$ and $\{1\}$. But $\pi(\{1\})$ is not open so, so $\{1\}$ cannot be open in $X$. By the definition of the subspace topology, it is not open in $X - \pi(0)$ either.

d) Consider the complement of $\pi([1/2])$ in $X$. Is it Hausdorff? Is it connected?

Yes it is Hausdorff. Since $X - \{0\}$ is open in $X$, $(X -\{0\}) \cap (X - \pi([1/2]) = \{1\}$ is open in the subspace topology. Similarly for $\{0\}$. Hence the subspace is Hausdorff.

It's not connected since $X - \pi([1/2]) = \{0\} \cup \{1\}$.

Are my answers correct?

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As I see, your answers are correct. There are minor issues with your definition of $H$. One thing is that $t$ stands there both for name of a function and a variable. Another is that you pass $t ∈ [0, 1]$ to $id_X$ but $X$ is not $[0, 1]$, there should be composition with the projection. But the idea is clear: you take the standard homotopy for contraction of $[0, 1]$ and show that it factors through $π$.

For other questions you may also forget $[0, 1]$ and see that $X$ is just three point space with a certain topology which can be easily visualized so the answers are clear.

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