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Let $f : X = [0,2] \cup (3,5] \to Y = [0,3]$ be defined by $$f(x) = \begin{cases} x &\text{ if } x \in [0,2] \\ x - 2 &\text{ if } x \in (3,5]. \end{cases}$$ I want to show that $f$ is a quotient map. It is easy to see that $f$ surjects and that $f$ is continuous, as it is linear on connected components of the domain. However, I am having trouble proving the forward direction of the quotient property, i.e., $f^{-1}(U)$ open in $X$ if and only if $U$ open in $Y$.

I also know a proposition that says if $q : A \to B$ is continuous and $\exists \ r: B \to A$ so that $r$ continuous and $q \circ r = id_B$, then $q$ is a quotient map. However, I am not sure if that quite helps. I came up with a function $r$, but it was not continuous.

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    If you look at the continuous extension $\tilde{f}\colon [0,2] \cup [3,5] \to Y$ of $f$, can you show that that is a quotient map? Can you from that deduce that $f$ is a quotient map? – 2017-01-06
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    I also know that if $q: A \to B$ is continuous, surjective and either open or closed, then it is a quotient map. That being said, my three approaches to this adjustment are a) show $\tilde{f}$ is open, b) closed, or c) use the property above. I think $\tilde{f}$ is not open, so should I try to show that it is closed? I don't see the proposition I gave as being useful for this function. – 2017-01-06
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    Seems like a good plan. What is a nice property of its domain that $X$ hasn't and that might be related to closedness of continuous maps? – 2017-01-06
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    Compactness and Hausdorff. Thank you. – 2017-01-06
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    On the other hand, it just dawned on me that it's not necessary to take that detour. A characterisation of the subsets $M$ of $Y$ such that $f^{-1}(M)$ is open, isn't actually difficult. – 2017-01-06
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    Oh? I assume we can write $f^{-1}(M)$ as a countable union of open intervals in the subspace topology, meaning we can reduce to looking at just open intervals since functions pushforward unions. – 2017-01-06
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    Way too complicated. A subset of $X$ is open if and only if ? and ?? are open? – 2017-01-06
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    If $O$ is our subset, then we have $O \cap [0,2]$ and $O \cap (3,5]$ open. Then use the fact that $f$ pushes forward unions and that $f$ restricted to connected components is open. – 2017-01-06
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    Yes. And if we translate that to $M\subset Y$, we have $f^{-1}(M)$ is open if and only if …? – 2017-01-06
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    $M$ open. Should I write out this out in the answer? How can I give you credit in the answer? – 2017-01-06

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$A \subset X$ is open if and only if $A\cap [0,2]$ and $A \cap (3,5]$ are both open. So for $U\subset Y$, $f^{-1}(U)$ is open if and only if $U \cap [0,2]$ and $U \cap (1,3]$ are both open (in the subspace topology). If $2\notin U$, then $U \cap [0,2] = U \cap [0,2)$, and so $U = \bigl(U \cap [0,2)\bigr) \cup \bigl(U \cap (1,3]\bigr)$ is the union of two open subsets of $Y$ if $f^{-1}(U)$ is open. And if $2 \in U$, then there is an $\varepsilon \in (0,1)$ such that $(2-\varepsilon, 2+\varepsilon) \subset U$, since $U \cap (1,3]$ is open, so $U = \bigl(U\setminus \{2\}\bigr) \cup (2-\varepsilon, 2+\varepsilon)$ is open because $U\setminus \{2\}$ is open in $Y$ by the previous part.