# A topological tree is weakly contractible

Let us call a nonempty topological space a topological tree if it is Hausdorff and for two distinct points there is a continuous injective path connecting the points, which is unique up to reparametrisation.

This should be equivalent to the following definition: The space $$X$$ is a topological tree if it is Hausdorff, pathwise connected and whenever $$\gamma_1,\gamma_2:[0,1]\to X$$ are topological embeddings with $$\gamma_1(0)=\gamma_2(0)$$ and $$\gamma_1(1)=\gamma_2(1)$$, it is true that $$\gamma_1([0,1])=\gamma_2([0,1])$$.

I am not reqiring the space to be paracompact nor to have the homotopy type of a CW-complex or anything like that, just a Hausdorff space.

Examples of such topological trees are $$\mathbb R$$, a hedgehog space (https://en.wikipedia.org/wiki/Hedgehog_space), or the Long Line (or the Long Ray).

What one could expect that such a space is always contractible as it is called a tree and trees should be contractible but the example of the Long Line shows that this is not the case.

However, it should be true that such a space is weakly contractible, i.e. every homotopy group is trivial. As $$\pi_0$$ is trivially trivial, I tried to show that for the fundamental group $$\pi_1$$ but was not able to do so.

Maybe this is a hard question like the one why two different points in a pathwise connected Hausdorff space can always be connected by an injective arc. I know where to find a proof for that fact, but is it frustrating that there seems to be not simple proof for that fact that seems obvious but is not.

By the way: I do not know if that defintion of a topological tree is already in use under a different name. I just took the notion of an $$\mathbb R$$-tree from metric geometry and removed the metric geometry out of the definition.

Another question would be if a topological tree is always 1-dimensional, using different notions of topological dimensions, but that should be a different question for a different day.

• It seems to me that your question reduces to the case of compact Hausdorff spaces, since the image of any map from a sphere to a topological tree should be a compact Hausdorff topological tree. It is clearly Hausdorff, path connected, and compact. That we can find nice paths follows from the answer to your "hard question" for existence, and the tree property of the whole space for uniqueness. Dec 19, 2022 at 9:30
• I am not sure that your two definitions are equivalent, unless I misunderstood your "pairwise connectedness": the first implies that the space is path-connected, while the sencond does not seem to be so, e.g. the topologist's sine curve.
– Z. M
Dec 19, 2022 at 10:14
• @Z.M The poster wrote pathwise connectedness, not pairwise connectedness. Dec 19, 2022 at 10:40
• @PierrePC: This is a very nice observation! Now that I think about it: It should even be possible to reduce the question to compact metrizable spaces as the continuous image of compact metrizable space inside a Hasdorff space is metrizable. Dec 19, 2022 at 13:04
• It seems that the alternative definition says your objects are real trees, see en.wikipedia.org/wiki/Real_tree Dec 20, 2022 at 11:51

Let $$X$$ be a a "topological tree" by your definition. Then $$X$$ is uniquely arcwise connected and Hausdorff. Let $$f:S^n\to X$$ be a map from the $$n$$-sphere where $$n\geq 1$$. It follows from the Hahn-Mazurkiewicz Theorem that the image $$f(S^n)$$ is a uniquely arcwise connected Peano continuum. This is equivalent to being a dendrite and dendrites are contractible. Thus $$f$$ contracts in $$f(S^n)$$ and it follows that $$\pi_n(X)$$ is trivial for all $$n\geq 0$$.