Universal property of quotient spaces
A typical textbook theorem about quotient space is as follows:
Theorem (Gamelin-Greene Introduction to Topology p.106): Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X/{\sim} \rightarrow Y$ such that $f=g \circ \pi$.
It is very easy to prove it by using the definition of continuous functions between topological spaces because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$. One such proof is done in the mentioned textbook as follows:
Proof. Since $f$ is constant on each equivalence class, one can define $g:X/{\sim}\to Y$ with $g([x])=f(x)$. One can easily check that $f=g \circ \pi$. To show that $g$ is continuous, consider any open set $V$ in $Y$. Since $f^{-1}(V)$ is open in $X$ by continuity of $f$ and $\pi^{-1}(g^{-1}(V))=(g\circ \pi)^{-1}(V)=f^{-1}(V)$, by the definition of the quotient topology on $X/{\sim}$, $g^{-1}(V)$ is open and the proof is complete.
But when I try to use the definition of continuity at a point, which I think is supposed to be easy too, I got stuck.
Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/{\sim}$ such that $g(O)\subset U$.
Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
Question: How can I go on from here or do I need to change the argument completely?
I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
2 answers
The issue is that for $g$ to be continuous it is not enough for $f$ to be continuous at $x$. It is not even enough for $f$ to be continuous at every $y \in [x]$, for example consider $f : \mathbf{R}_{>0} \to \{0,1\}$ which takes $(n- 1/n, n + 1/n)$ to $1$ and every other point to $0$ and the equivalence $\sim$ which identifies $\mathbf{N}_{> 0} \subset \mathbf{R}_{> 0}$, ie $x \sim y$ iff $x = y$ or $x, y \in \mathbf{N}$. Then $g$ takes $[1]$ to $1$, but every neighborhood of $[1]$ contains a point where $g$ takes the value $0$.
So you need to use continuity of $f$ "uniformly" around all points of $[x]$ simultaneously, and the correct notion of "uniform" is to consider $\pi^{-1}(V)$ for neighborhoods $V$ of $[x]$.
If the quotient map $\pi$ happens to be open, for instance if $\sim$ is induced by a continuous action of a group, then it is enough to know that $f$ is continuous at each $y \in [x]$: take $V_y \subset f^{-1}(U)$ a neighborhood of $y$, then $\pi(\bigcup V_y)$ is a neighborhood of $[x]$ contained in $g^{-1}(U)$.
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I figured out an answer after posting the question for a while. I would like to record it here.
The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$. One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$, which, of course, could have been used in the simple proof via continuity of functions between topological spaces.
Proof. Define $g:X/{\sim}$ as in the standard proof. Without loss of generality, suppose $U$ is an open neighborhood of $g([x])$. Then $\pi^{-1}(g^{-1}(U))=f^{-1}(U)$ is open by the continuity of $f$ and thus $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ by the definition of quotient topology. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$. Since $[x]$ is arbitrary, the proof is complete.
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