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.



A topological tree is weakly contractible
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](https://wildtopology.com/2019/08/06/shape-injectivity-of-the-hawaiian-earring-part-ii/). Thus $f$ contracts in $f(S^n)$ and it follows that $\pi\_n(X)$ is trivial for all $n\geq 0$.


The results I'm using here are part of "Continuum Theory." Nadler's book has a great chapter on dendrites.


