I am doing a self-review for the test in topology, and I was looking for proof verification of the following problem.
Problem: Prove that function defined by $$f:X=[0,1]\cup[2,3]\rightarrow Y=[0,2]:\\[1em]x\mapsto \begin{cases} x,&x\in[0,1]\\
x\mapsto x-1, &x\in[2,3]\end{cases}$$
is a quotient map, where domain and range have subspace topology inherited from Euclidean topology on the real line.
Proof: The idea is to prove three facts that would verify by definition that the function is a quotient map:
1. the function is continuous 2. the function is surjective 3. if preimage of a set is open then the set is open.
The first two propositions are easily verified (1-by pasting lemma of two continuous functions, 2-immediately observed). However, the third part is a little tricky.
For 3rd proposition, we need that given $U\subset Y$, if $f^{-1}(U)$ is open, then $U$ is open. I intend to prove this by proving that for each $y\in U$, there exists an open set that contains $y$ and is contained in $U$. Consider cases (note that due to surjectivity: $f(f^{-1}(y))=y$):
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$y=2$. Then, $f^{-1}(y)=\{3\}$. Since $f^{-1}(U)$ is open, hence there exists an open set of the form $(3-\delta,3]\subset [2,3]$, such that $f^{-1}(y)\in(3-\delta,3]\subset f^{-1}(U)$. Therefore, $y=2\in f((3-\delta,3])\subset f(f^{-1}(U)) \subset U$. But $f((3-\delta,3])=(2-\delta,2]$ is open in $Y$. Hence, y is contained in the open that is contained in $U$. Similarly, we can treat the case when $y=0$.
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$y\in (1,2)$. Then, $f^{-1}(y)\in (2,3)$. Since $f^{-1}(U)$ is open, hence there exists an open set of the form $(f^{-1}(y)-\delta,f^{-1}(y)+\delta)\subset (2,3)$, such that $f^{-1}(y)\in (f^{-1}(y)-\delta,f^{-1}(y)+\delta)\subset f^{-1}(U)$. Therefore, $y\in f((f^{-1}(y)-\delta,f^{-1}(y)+\delta))\subset f(f^{-1}(U)) \subset U$. But $f((f^{-1}(y)-\delta,f^{-1}(y)+\delta))=(f^{-1}(y)-\delta-1,f^{-1}(y)+\delta-1)$ is open in $Y$. Hence, y is contained in the open that is contained in $U$. Similarly, we can treat the case when $y\in (0,1)$.
- $y=1$. Then, $f^{-1}(y)=\{1,2\}$. Since $f^{-1}(U)$ is open, hence there exists an open set of the form $(1-\delta,1]\cup [2,2+\delta)\subset X$, such that $f^{-1}(y)\subset (1-\delta,1]\cup [2,2+\delta)\subset f^{-1}(U)$. Therefore, $y\in f((1-\delta,1]\cup [2,2+\delta))\subset f(f^{-1}(U)) \subset U$. But $f((1-\delta,1]\cup [2,2+\delta))=(1-\delta,1+\delta)$ is open in $Y$. Hence, y is contained in the open that is contained in $U$.
Therefore, $U$ is open in Y.
Best Answer
Your proof is correct. However, you can do it simpler.
First observe that the restrictions $f_1 : [0,1) \to [0,2]$ and $f_2 : (2,3] \to [0,2]$ of $f$ are open maps, i.e. send open sets to open sets.
Let $U \subset [0,2]$ be a set such that $V = f^{-1}(U)$ is open. We can write $V = V_1 \cup V_2$ with open $V_1 = V \cap [0,1] \subset [0,1], V_2 = V \cap [2,3] \subset [2,3]$. Let $V'_1 = V_1 \cap [0,1)$ which is open in $[0,1)$ and $V'_2 = V_2 \cap (2,3]$ which is open in $(2,3]$. Then $f(V'_1 \cup V'_2) = f_1(V'_1) \cup f_2(V'_2)$ is open in $[0,2]$.
Case 1. $1 \notin U$. Then $1 \notin V_1, 2 \notin V_2$. Hence $V'_1 = V_1$ and $V'_2 = V_2$. Thus $V = V'_1 \cup V'_2$ and $U = f(V'_1 \cup V'_2)$ is open.
Case 2. $1 \in U$. Then $1 \in V_1, 2 \in V_2$. There is $\epsilon > 0$ such that $(1-\epsilon,1] \subset V_1, [2,2+\epsilon) \subset V_2$. Hence $V = V'_1 \cup V'_2 \cup(1-\epsilon,1] \cup [2,2+\epsilon)$ and $U = f(V) = f(V'_1 \cup V'_2) \cup f((1-\epsilon,1]) \cup f([2,2+\epsilon)) = f(V'_1 \cup V'_2) \cup (1-\epsilon,1+\epsilon)$ is open.
Edit:
Even simpler: The restrictions $f_1 : [0,1] \to [0,2]$ and $f_2 : [2,3] \to [0,2]$ of $f$ are closed maps. Let $A \subset [0,2]$ be a set such that $B = f^{-1}(A)$ is closed. We can write $B = B_1 \cup B_2$ with closed $B_1 = B \cap [0,1] \subset [0,1], B_2 = B \cap [2,3] \subset [2,3]$. Then $A = f(B) = f(B_1 \cup B_2) = f(B_1) \cup f(B_2) = f_1(B_1) \cup f_2(B_2)$ is closed.