Let $\omega_1$ be the first uncountable ordinal. The long line $L$ is defined as the cartesian product of the first uncountable ordinal $ \omega _{1}$ with the half-open interval $ [0,1)$ equipped with the order topology that arises from the lexicographical order on $\omega _{1} \times [0,1)$. That means that of we set by $"<_L"$ the lexicographic order on $\omega_1 \times [0,1)$ then the topology on $\omega_1 \times [0,1)$ is generated by "interval" base
$$ ((k_1,a), (k_2,b))=\{(k,x) \in \omega_1 \times [0,1) \vert \ \vert (k_1,a) <_L (k,x) <_L (k_2,b)$$
My question is if it's true that the fundamental group $\pi(L,l_0)$ of the long line is trivial (let $l_0$ is any base point) or what's the reason? Although it seems to be a rather natural question I nowhere found a serious discussion about this problem. My intuition says that it should be trivial, but I'm not not sure.
My idea: I have to show that any loop $f: S^1 \to L$ is contractible. My idea is to observe that one can find a finite number of "neighboured" $a_1, a_2,..., a_n \in \omega_1$ and show that the image of $f$ is contained in $\bigcup_{i=1}^n {a_i} \times [0,1) \subset L$ and then because of finiteness of $a_1, a_2,..., a_n \in \omega_1$ one sees that $\bigcup_{i=1}^n \{a_i\} \times [0,1) \cong [0,1)$ and this space is contractible.
Is the argument ok? My worries are based on my doubts if that's make sense to say that we can fetch some "neighboured" $a_1, a_2,..., a_n \in \omega_1$. What should "neighboured" here mean? If we believe in continuum hypothesis then $\omega_1= \mathbb{R}$ and it is strage to say what does it mean that two elements $a, b \in \mathbb{R}$ a "neighboured". If $a \neq b$ there is always some other element between them, therefore I doubt that my argument above should work, or not?
note that we were done if we could show that every loop $f: S^1 \to L$ must be contained in a subset of the form $\{r\} \times [0,1) \subset L$. This one is clearly contractible, so the loop would be also contractible. Is it possible to do it that way around?
Does somebody see how can I argue to conclude that the fundamental group of $L$ is trivial (if that's really the case)?
The key to understanding questions like this about the long line is the following observation.
Lemma: For any $x\in L$, the interval $I_x=\{y\in L:y\leq x\}$ is order-isomorphic to $[0,1]$.
Proof: Let $A$ be the set of $y\in L$ such that $(0,0)<y<x$ and the second coordinate of $y$ is rational. Then $A$ is countable (since the first coordinate of $x$ is a countable ordinal so there are only countably many choices for the first coordinate of $y$) and densely ordered without endpoints, so it is order-isomorphic to $(0,1)\cap\mathbb{Q}$. Also, $A$ is dense in $I_x$ and $I_x$ is Dedekind-complete (and has endpoints), so $I_x$ is the Dedekind-completion of $A$ (including endpoints). So, $I_x$ is order-isomorphic to the Dedekind completion (including endpoints) of $(0,1)\cap\mathbb{Q}$, which is just $[0,1]$. $\blacksquare$
This Lemma is really the reason that $L$ is called the "long line". If you cut it off at any point before the end, it really does just look like the ordinary real line. To see that it is different, you have to go all the way "uncountably far" to the end.
In particular, it follows that $I_x$ is homeomorphic to $[0,1]$ and contractible. But any continuous map from a separable space to $L$ has image contained in $I_x$ for some $x\in L$ (take $x$ to be an upper bound for the image of a countable dense subspace). So, any continuous map from a separable space to $L$ is nullhomotopic. In particular, this implies all the homotopy groups of $L$ are trivial.
(Note that the Lemma also implies that the image of a loop in $L$ does not have to be covered by finitely many sets of the form $\{a\}\times[0,1)$. Indeed, for any infinite ordinal $\alpha<\omega_1$, the interval $I_{(\alpha,0)}$ contains infinitely many sets of the form $\{a\}\times[0,1)$, but can be covered by a single loop since it is homeomorphic to $[0,1]$.)