# Proper maps for connectedness

A map $f : X \to Y$ between topological spaces $X$ and $Y$ is said to be proper if the inverse image of a compact subset of $Y$ is compact in $X$.

Does there exists an analogous concept for connectedness?

A map $f : X \to Y$ between topological spaces $X$ and $Y$ is said to be ???? if the inverse image of a connected subset of $Y$ is connected in $X$.

If such a concept does not exist or hasn't been studied much, is there a reason for this? Why in that case has compactness seemingly been deemed of more important than connectedness?

• I would think it's a very rare property. Something as innocuous as $f(x)=x^2$ on the reals fails it: $f^{-1}(1,+\infty)$ is disconnected. – Randall Sep 22 '17 at 11:56
• Yes, there is such a concept; it is useful in algebraic topology, start by reading here. – Moishe Kohan Sep 23 '17 at 5:29

## 1 Answer

Sometimes a map between topological spaces $f\colon X \to Y$ is called monotone if it has connected fibers, i.e. $f^{-1}(y)$ is connected for every $y ∈ Y$. (Similarly, $f$ is called perfect if it has compact fibers and it is closed and continuous. Such maps are proper.)

If $f$ is a monotone quotient map and $Y$ is connected, then $X$ is connected as well. Hence, if $f$ is monotone and hereditarily quotient, then the preimage of a connected set is connected. (Hereditarily quotient means that every corestriction $f\colon f^{-1}[B] \to B$ is quotient. Open or closed continuous maps are such.)

• Great answer! Just to make sure, in this paper, in the second paragraph after Proposition 2.2, the third sentence reads: "Since the inverse image, under a monotone quotient map, of a closed connected set is connected...". Since monotone maps have connected (particularly non-empty) fibers they are surjective, so a monotone quotient map is open and so it seems the emphasized closed may be omitted entirely. Is this correct? – Arrow Dec 13 '18 at 19:15
• @Arrow: I don't think so. Monotone map does not have to be onto, the empty set is also connected. On the other hand, quotient map is always onto. But it is not true that a monotone quotient map is open, so your map is not necessarily hereditarily quotient. But for closed or open set, the corresponding corestriction of a quotient map is quotient. So the closedness is used in the argument. – user87690 Dec 13 '18 at 21:55
• For many people (me included!) connected implies non-empty :) – Arrow Dec 13 '18 at 22:09
• @Arrow: Sure, that's a matter of taste. But a monotone map does not have to be onto, e.g. $\operatorname{arctan}\colon ℝ \to ℝ$ is monotone. – user87690 Dec 13 '18 at 22:27
• Dear @user87690, on a very related manner, could you please take a look at Engelking 6.2.23? The mapping $h(z)=g_2g_1^{-1}(z)$ seems to be undefined if $g_1^{-1}(z)$ is empty, and this seems to invalidate the proof. So it appears having (nonempty) connected fibers is crucial for essential uniqueness of monotone-light factorizations. What do you think? – Arrow Dec 14 '18 at 11:17