Let $X$ be a $\Delta$-generated space having a subset $A=\{a,b\}$ such that the relative topology is the Sierpinski topology with for example $\{a\}$ closed and $\{b\}$ open (the Sierpinsky space is a $\Delta$-generated space). I consider the quotient set map $X\to X / (a=b)$ with $X / (a=b)$ equipped with the final topology.
I want to prove that $X\to X / (a=b)$ is a homotopy equivalence. I do not know if I am clumsy or if the statement is just wrong.
The idea I am exploring is to start from the proof that the Sierpinski space is contractible. Let $H:[0,1] \times A \to A$ be the set map defined by $H(0,a)=a$, $H(0,b)=b$, $H(1,a)=H(1,b)=a$ and $H(u,a)=H(u,b)=b$ for $u\in ]0,1[$. Then $$H^{-1}(b) = \big(]0,1[ \times \{a,b\}\big) \cup \big(\{0\} \times \{b\}\big).$$ Since $]0,1[ \times \{b\} \subset ]0,1[ \times \{a,b\}$, we deduce that $$H^{-1}(b) = \big(]0,1[ \times \{a,b\}\big) \cup \big([0,1[ \times \{b\}\big)$$ which is open. Therefore $H$ is continuous and it is a homotopy between the identity of $A$ and the constant map $a$. The statement above is therefore true when $X=A$.
Then I extend $H$ to a set map $H:[0,1]\times X \to X$ by setting $$\forall x\in X\backslash A, \forall u\in [0,1], H(u,x)=x.$$ One has $$H^{-1}(b) = \big(]0,1[ \times \{a,b\}\big) \cup \big([0,1[ \times \{b\}\big)\\H^{-1}(a) = \big(\{1\} \times \{a,b\}\big) \cup \big(\{0\} \times \{a\}\big) \\ H^{-1}(A) = [0,1] \times A \\ \forall x\in X\backslash A, H^{-1}(x) = [0,1]\times \{x\} . $$ I need to prove that $H:[0,1]\times X \to X$ is continuous. If $U$ is an open of $X$, there are three mutually exclusive cases:
- $U\cap A =\varnothing$. Then $H^{-1}(U)=[0,1]\times U$ which is an open of $[0,1] \times X$.
- $A\subset U$. Then $H^{-1}(U)=[0,1]\times U$ as well which is an open of $[0,1] \times X$.
- $b\in U$ and $a \notin U$. Then $$H^{-1}(U)=H^{-1}(U\backslash \{b\}) \cup H^{-1}(b).$$ We obtain $$H^{-1}(U)=\big([0,1]\times (U\backslash \{b\})\big) \cup \big(]0,1[ \times \{a,b\}\big) \cup \big([0,1[ \times \{b\}\big).$$ Since $[0,1[\times (U\backslash \{b\}) \subset [0,1]\times (U\backslash \{b\})$, we obtain $$H^{-1}(U)=\big(\{1\}\times (U\backslash \{b\})\big) \cup \big(]0,1[ \times \{a,b\}\big) \cup \big([0,1[ \times U\big).$$ Since $b\in U$ by hypothesis, we obtain $$H^{-1}(U)=\big(\{1\}\times (U\backslash \{b\})\big) \cup \big(]0,1[ \times \{a\}\big) \cup \big([0,1[ \times U\big).$$ I cannot prove that $H^{-1}(U)$ is open...
