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Made question better defined and clarified the closed disk.
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Isky Mathews
Isky Mathews
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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 (where implicitly we're embedding the tree in $\mathbb{R}^2$ already).

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 $\mathbb{S}^1$).

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 is a retraction of the disk (where implicitly we're embedding the tree in $\mathbb{R}^2$ already).

Intuitively, each such tree is not only 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.

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 $\mathbb{S}^1$).

Source Link
Isky Mathews
Isky Mathews
  • 91
  • 8

Is every tree a deformation retract of the disk?

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 is a retraction of the disk (where implicitly we're embedding the tree in $\mathbb{R}^2$ already).

Intuitively, each such tree is not only 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.