Every continuous function from $\mathbb{R}$ to $\mathbb{R}$ is a quotient map

Question

A continuous surjection $f:X\to Y$ is a quotient map if the topology on $Y$ is the final topology with respect to $f$, namely, for each subset $F\subset Y$, $F$ is closed if $f^{-1}(F)$ is closed in $X$.

Equivalently, for each saturated closed subset $B$ of $X$, $f(B)$ is closed in $Y$. (Here, saturated = saturated with respect to $f$, i.e., saturated with respect to the fibers $f^{-1}(y)$.)

All the examples of a map that is not quotient that I could find in various references or on this site were always a little complicated. So I tried to find a simple example of a continuous map from the reals to the reals that is not a quotient map. After a while I realized it is not possible:

Every continuous map $f:\mathbb{R}\to\mathbb{R}$ is a quotient map, when viewed a surjection onto its image.

Apart from checking that my reasoning below is correct, can you provide a different proof? Specifically, is this a consequence of some more general topological results?

---

Proof of result: Let $X=Y=\mathbb{R}$ and $f$ be a continuous map from $X$ to $Y$. Let $B$ be a saturated closed set in $X$. We have to show that $f(B)$ is closed in $f(X)$. Suppose that is not the case. So we can find $a\in X$ with $f(a)\notin f(B)$, but such that $f(a)$ is a limit point of $f(B)$. There is a sequence $(x_n)$ in $B$ such that $f(x_n)$ converges to $f(a)$. Taking a subsequence if necessary, we can assume that $f(x_n)$ converges monotonically to $f(a)$. Without loss of generality, say it's an increasing sequence, and $a<x_1$.

We have $f(x_1)<f(x_n)<f(a)$, so by the intermediate value theorem there is a $z_n\in[a,x_1]$ such that $f(z_n)=f(x_n)$. Since $B$ is saturated, the point $z_n$ also belongs to $B$. Thus $(z_n)$ is a bounded sequence in $B$ and by compactness there is a subsequence converging to a point $z\in B$ (because $B$ is closed). Replacing $(z_n)$ again by that subsequence, we can assume $z_n$ converges to $z$, and by continuity of $f$ we have $f(z_n)$ converging to $f(z)$. But that is the same as $f(x_n)$, which converges to $f(a)$. So $f(a)=f(z)$ did belong to $f(B)$ after all.

Answer 1

Here is an argument that uses the same idea of your proof but doesn't involve sequences.

If $X$ is path connected and $Y$ is a totally ordered set with order topology, then every continuous map $f:X\to Y$, viewed as a surjection onto its image, is a quotient map.

Proof: First note that $f(Y)$ is connected thus must convex, thus the order topology and subspace topology coincide, so it suffices to consider a surjective $f:X\to Y$. If $B$ is saturated closed, we show that $f(B)$ is also closed. For every $[y,y']\subset Y$, choose $x,x'$ with $f(x)=y,f(x')=y'$, and let $P$ be a path connecting $x,x'$, then $[y,y']\subset f(P)$ by intermediate value theorem(or just by connectness). Note that $B$ is saturated, so $f(P)\cap f(B)=f(P\cap B)$ is compact, and $[y,y']\cap f(B)$ is also compact hence closed. Since every space is coherent with any family of subsets whose interiors cover it Thus all intersections $]y,y'[\cap f(B)$ (or $[a,y'[\cap f(B),]y,b]\cap f(B)$ where $a,b$ are minimal or maximal in $Y$ if they exist) are also closed in $]y,y'[$ (or $[a,y'[, ]y,b]$ respectively). Since these intervals form a basis of $Y$, we are done.

Rmk1: This cannot be generalised to connected $X$: $\mathbb R_K$ is connected, but $\mathrm{id}:\mathbb R_K\to\mathbb R$ is continuous but not a quotient map.

Rmk2: As a consequence, all topology on an ordered set that strictly finer than the order topology is not path connected.