4
$\begingroup$

I apologise if this question is not suitable for MathOverflow.

We define a graph here to mean a disjoint union of points and copies of $[0,1]$ quotiented so that the endpoints of any interval lie on a point (in particular, a graph is a topological space and we permit loops and multiple edges going between vertices). My friend and I have been taking an algebraic topology course at our university recently and were trying to classify exactly the finite graphs with the Fixed Point Property. We were able to show via reasoning relating to edge contractions yielding retractions of graphs that any non-simple graph doesn't have the Fixed Point Property and then later that any simple graph with a cycle fails to have FPP.

Thus we reduced our reasoning to the case of trees. Now, we have an argument where you use the fact that the disk has no retraction to the circle to show that any tree has the FPP but it requires the fact that any such tree has an embedding into the disk so that it is a retraction of the disk.

Intuitively, each such tree is not only embeddable as a retract but a deformation retract, but we are not sure how to prove such a general result (given we only have learnt basic results). How might you go about proving this? Is the same result true for infinite trees?

Answers are permitted to use various tools possibly more advanced than my current knowledge-level as I'll be happy to read up on these other ideas/techniques.

EDIT: Changed the question to look for an embedding into the disk so that it is a deformation retract, as my original question was poorly defined as pointed out in the comments. Also to be clear, we are discussing the closed disk (as otherwise we couldn't even discuss retractions from the disk to $\mathbb{S}^1$).

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ One easy way to do this (for finite trees) is to give your tree the structure of a fatgraph. To do this, you put a cyclic ordering on the edges incident at each vertex, and there's also an obvious orientation condition on each edge. It's easy to prove by induction that any finite graph has the structure of a fatgraph. It's also easy to see that a fatgraph structure allows you to "thicken" your graph $G$ to a compact surface with boundary $S(G)$, in such a way that $S(G)$ deformation retracts to $G$... $\endgroup$
    – HJRW
    Commented Nov 22, 2022 at 12:07
  • 2
    $\begingroup$ (cont'd from above.)... Finally, if $T$ is a finite tree then $S(T)$ is a compact, connected surface of Euler characteristic 1, so is homeomorphic to a disc by the classification of surfaces. $\endgroup$
    – HJRW
    Commented Nov 22, 2022 at 12:08
  • 1
    $\begingroup$ Are you allowing the "long line" as a tree? $\endgroup$
    – Jonny Evans
    Commented Nov 22, 2022 at 12:35
  • 1
    $\begingroup$ @HJRW: Yeah indeed! I was in a lecture during the asking of this question and while I was away typing things up I realised this quickly and came back to mention it but I guess it's very obvious, as you say. $\endgroup$
    – Isky Mathews
    Commented Nov 22, 2022 at 12:46
  • 2
    $\begingroup$ Just a remark: any graph embedding in the disk (open or closed, or any manifold for that matter) has to be at most countable. Indeed the set of vertices is discrete in the graph topology so it must be discrete, hence at least countable, in the subset topology as well. $\endgroup$
    – Pierre PC
    Commented Nov 22, 2022 at 12:56

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .