# Universal property of quotient spaces

A typical textbook theorem about quotient space is as follows:

Theorem(Gamelin-GreeneIntroduction to Topologyp.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.

## 1 answer

^{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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