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

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.

We have $$f(x_1), so by the intermediate value theorem there is a $$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.

• Yes, this proof is correct. Nice observation! The simplest commonly used example of a surjective continuous map which is not a quotient map is not at all complicated; it is $f: [0, 2\pi)\to S^1\subset \mathbb C$, $f(t)=e^{it}$. Commented Jul 27 at 23:47
• @PatrickR. I think need $Y$ be 1st countable to use sequences(I'm not sure if every ordered space is 1st countable), but I think this can be avoided. Commented Jul 31 at 2:24

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.

• By "ordered space", I assume you mean what is usually called a LOTS (linearly ordered topological space): topology.pi-base.org/properties/P000133. Commented Jul 31 at 4:22
• Yes. I will add some clarification. Commented Jul 31 at 5:02
• Nice generalization. Note that if $Y$ is a LOTS and $A$ is a connected subset of $Y$, the subspace topology on $A$ coincides with the order topology induced by the restriction of the order to $A$. In brief, a connected subspace of a LOTS is also a LOTS. So $f(X)$ is also a LOTS and there is no loss of generality in just assuming $f$ is surjective, like you originally had it. Either way is fine. Commented Jul 31 at 5:11
• Very nicely explained! Commented Jul 31 at 5:16