# Snake-like continua and universal images

Among the Hausdorff compact spaces the closed interval is the simplest snake-like continuum. I'll present the definition after stating the problem.

The snake-like continua $\ S\$ are universal images for Hausdorff compact spaces in the following sense:

THEOREM   Let $\ S\$ be an arbitrary snake-like continuum. Let $\ u:X\rightarrow S$ and $\ f:X\rightarrow S\$ be two arbitrary continuous maps of an arbitrary connected Hausdorff space $\ X\$ into $\ S\$ such that $\ u\$ is a surjection. Then there exists $\ p\in X\$ such that $\ f(p)=u(p)$.

The questions is: are the snake-like continua the only universal images for the Hausdorff connected spaces? Let me state it in detail:

QUESTION   Let an arbitrary Hausdorff compact connected space $\ Y\$ fail to be snake-like. Does there exist a Hausdorff connected compact space $\ X\$ and two continuous maps $\ v:X\to Y$ and $\ f:X\rightarrow Y\$ such that $\ v\$ is a surjection, and there does not exist any $\ p\in X\$ such that $\ f(p)=v(p)$.

D   E   F   I   N   I   T   I   O   N(s)

Let $\ U\subseteq X\times X.\$ A function $\ f:X\rightarrow Y\$ is called a $U$-function $\ \Leftarrow:\Rightarrow\ \forall_{y\in Y}\ f^{-1}(y)\times f^{-1}(y) \subseteq U$.

Thus $\ \Delta_X:=\{(x\ x) : x\in X\}\ \subseteq U\$ for every $U$-function $\ f:X\rightarrow Y\$ which is a surjection.

Let $\ \mathbf U:= \mathbf U_U\$ stand for all $U$-functions.

DEFiNITION   A Hausdorff compact connected space $\ S\$, with the induced topology $\ T\$ in $\ S\times S,\$is called snake-like $\ \Leftarrow:\Rightarrow$

$$\forall_{U\in T}\ \left(\ \Delta_S\subseteq U\ \Rightarrow\ \exists\ \left(f:S\rightarrow[0;1]\right)\in \mathbf U\ \right)$$

The metric compact case

Let $\ (X\ d)\$ be a compact metric space, and let $\ \epsilon>0.\$ Let

$$U(\epsilon)\ :=\ \{(x\ y)\ \in\ X\times X\ :\ d(x\ y)<\epsilon\ \}$$

Then $\epsilon$-maps are defined as $\mathbf U$-maps, where $\ \mathbf U:=\mathbf U_{U(\epsilon)}.\$ Thus it's quite obvious that in the case of compact metric spaces we can talk about $\epsilon$-maps instead of $\mathbf U$-maps, it's equivalent--indeed, family $\ \{U(\epsilon) : \epsilon > 0\}\$ is a base of open neighborhoods of the diagonal $\ \Delta_X\$ in $\ X\times X$.

• I am a bit confused by the type-setting. Is that what is supposed to be in the grey box? Apr 12, 2015 at 19:43
• Confused?? There are no special rules about the contents of the gray fields that I know of. Apr 13, 2015 at 3:00
• @DavidWhite It's there by intention, as several recent discussions have borne out. Probably no one but the author understands why he type-sets that way, but I think it would be wise not to pursue this here and now. The mathematics looks interesting. Apr 13, 2015 at 4:10
• @WłodzimierzHolsztyński I hope you don't find my edits intrusive; I made them because I was genuinely confused about the $u\ f$ and $v\ f$ notation. I apologize if they constitute an overreach, I know you are sometimes particular about wording. Apr 13, 2015 at 8:25
• Actually we don't have to know what snake-like is: if I understand correctly we just need to answer this question: Suppose $X,Y$ are compact Hausdorff and connected, and $f,v:X\to Y$ are continous with $v$ surjective. Then: is there always $p\in X$ sucht that $f(p) = v(p)$? Apr 13, 2015 at 8:37

EDIT. As requested, I am extending the answer, including the relevant definitions. To do so, I have also re-arranged the answer somewhat. I am also now including the notions of surjective span and semi-span, which are directly relevant. As before, all continua in the following are assumed metrisable.

Summary. Your question is directly related to some long-standing notions and questions in continuum theory, due to Lelek. Your condition of being a "universal image" is precisely the same as having "surjective semispan zero". This condition in turn is satisfied by any continuum having span zero. These notions were introduced by Lelek to study arc-like (=snake-like) continua. For a long time, it was conjectured that every span zero continuum is arc-like, but this was disproved by Hoehn in 2011 (A non-chainable continuum of span zero, Fund. Math. 211 (2011), DOI:10.4064/fm211-2-3). So this work gives a counterexample to your question also.

Span and semispan. A continuum $X$ has span zero, abbreviated as $\sigma(X)=0$, if every continuum $C\subset X^2$ with

1. $\pi_1(C)=\pi_2(C)$

must intersect the diagonal; i.e. there is $x\in X$ with $(x,x)\in C$.

If we replace 1. by one of the following:

1. $\pi_1(C)\supseteq \pi_2(C)$;
2. $X=\pi_1(C)=\pi_2(C)$;
3. $X=\pi_1(C)$;

then we say that $X$ has

1. semispan zero ("$\sigma_0(X)=0$");
2. surjective span zero ("$\sigma^*(X)=0$");
3. surjective semispan zero ("$\sigma_0^*(X)=0$").

Remark. Let $X$ be a metric continuum, or even more generally a connected metric space. Then the "span" $\sigma(X)\in [0,\infty]$ is defined as the supremum over $\inf_{(a,b)\in C} d(a,b)$ for continua $C$ as above; similarly for $\sigma_0$ etc. However, we only require the definition of zero span; note that the latter is a purely topological notion.

All of these conditions are due to Lelek (see the references of the papers cited below). The rough idea of the definition is that, in a span zero continuum, two points cannot exchange position without meeting somewhere.

The following implications are known: $$X\text{ arclike}\quad \Rightarrow \quad \sigma_0(X)=0 \quad \Leftrightarrow \quad \sigma(X)=0 \quad \Rightarrow \quad \sigma_0^*(X)=0 \quad\Rightarrow\quad \sigma^*(X)=0.$$ Here the first implication is due to Lelek. The equivalence of zero span and semispan is due to Davis (The equivalence of zero span and zero semispan, Proc. AMS 90 (1984), DOI:10.1090/S0002-9939-1984-0722431-3). Given the latter result, the other implications are obvious.

Equivalent formulations. Observe that these conditions can be reformulated using a pair of continuous functions on a continuum, in the spirit of your question as follows. I believe this observation was also already made by Lelek. The relevant one for us is as follows.

Observation. A continuum $X$ has surjective semispan zero if and only if, for any continuum $Y$ and any continuous functions $v,f\colon Y\to X$ with $v(Y)=X$, there is $y\in Y$ such that $v(y)=f(y)$.

Proof. Assume $\sigma_0^*(X)= 0$, and let $Y,v,f$ be as above. Then we set $$C:= \{(v(y),f(y))\colon y\in Y\}.$$ Then $C$ satisfies 4. above, and hence intersects the diagonal.

For the opposite direction, we take $Y:= C$, $v:= \pi_1$ and $u:= \pi_2$. $\blacksquare$

So your question is equivalent to the following:

Question. Does $\sigma_0^*(X)=0$ imply that $X$ is arc-like?

The following two questions are due to Lelek:

Question 1. Does $\sigma(X)=0$ imply that $X$ is arc-like?

Question 2. Does $\sigma_0^*(X)=0$ imply $\sigma_0(X)=0$?

(Actually, Lelek asked a stronger question than Question 2 about connected metric spaces.)

Question 1 was particularly famous, and appeared in a number of problem lists. As mentioned above, it was disproved by Hoehn in 2011, giving a negative answer to your question.

I am not sure about the state of Question 2 for continua (but please note that I am not an expert in the area). A search did find an article (Ye, Liu, A space with infinite span and zero surjective span, Topology and its Applications 2001, DOI:10.1016/S0166-8641(00)00045-6), answering a related question of Lelek, but for non-compact spaces. I haven't checked this in detail, but from the summary it does not seem to lead easily to an answer for continua.

EDIT. Logan Hoehn kindly confirmed that the question of span versus surjective span is still open, as far as he knows. In his thesis, he gave an example of a continuum whose span equals four times the surjective span. (Lelek asked whether the span is always at most twice the surjective span.) This still seems to be the best-known result.

• My general comment addressed to the whole MO: let's make MO posts more self-contained, let them include explicit definitions. Apr 30, 2016 at 2:08
• The concepts of span and semispan - which are classical and well established, though admittedly somewhat specialised - can be found in the references provided. I think for an answer, as opposed to a question, that should be sufficient, but I will add some details when I get the chance. Apr 30, 2016 at 13:38
• I'd appreciate it. People are different :-). I read the respective definitions a few times in my life (years ago). And I worked on somewhat related problems. Nevertheless, I have never mastered those definitions. They are perhaps even less known to non-specialists while these topics have a universal appeal (I'd think). Let's remember the remarkable story of Paul Erdos and his under-additive 1-dim example. Apr 30, 2016 at 13:50
• I enjoy thinking about these concepts and problems, so I have added a detailed explanation - I hope this helps. However, in general, I feel that - while more detailed answers are always nice, and likely to help more people - it is enough if a MO answer gives the relevant information and references to definitions and further background. Certainly if I think a question is worth asking on MO, I also think it is worth the effort of reading up on any material mentioned in answers. May 3, 2016 at 11:01
• Speaking of self contained, would you might retelling the Erdos story here? May 3, 2016 at 12:27