In the simplest cases, the fundamental group serves as a measure of the number of 2-dimensional "holes" in a space. It is interesting to know whether they capture the following type of "hole".
This example may look pathological, but one must understand where one gets stuck, when one tries to study pathological spaces. It helps one in understanding where exactly all the extra nice conditions are used, and hopefully this type of approach will help in minimizing the number of false beliefs we unconsciously have.
The line with the double origin is the following space. In the union $\{0\} \times \mathbb R \cup \{ 1 \} \times \mathbb R$, impose the equivalence relation $(0, x) \sim (1, x) $ iff $x \neq 0$.
This space is locally like the real line, ie a $1$-manifold in everything except the Hausdorff condition. It is connected, path-connected and semilocally simply connected. Just the sort of nice space you study in the theory of fundamental group and covering spaces, except for the (significant) pathology that it has one inconvenient extra point violating the Hausdorff-ness.
It seems that the usual methods of computing the fundamental group are not working for this space. Van Kampen's theorem in particular does not apply. Also the covering spaces are weird, just like this space. In fact this space would have been a covering of $\mathbb R$, were it not for the condition that the preimage of every point is a disjoint union of open sets.
So, what if we try to compute the fundamental group of this space? I would be satisfied to know whether it trivial or not. Say, is the collection of homotopy classes of loops based at $1$ nontrivial? It is possible to speculate that a certain loop based at $1$ which passes through both the origins on this special line, in such a way that it passes through the "upper" origin, ie $(1,0)$ on the way left, and it passes through the lower origin, ie $(0,0)$, ought not to be homotopic to the constant loop based at $1$. But how to go about proving/disproving this statement?
Best Answer
The earlier answers showing that the fundamental group of this space is infinite cyclic by determining its universal cover or by constructing a fiber bundle over it with contractible fibers are very nice, but it's also possible to compute $\pi_1(X)$ by applying the classical van Kampen theorem not to $X$ itself but to the mapping cylinder of a map from the circle to $X$ representing the supposed generator of $\pi_1(X)$, namely the map that sends the upper and lower halves of $S^1$ to arcs in $X$ from $+1$ to $-1$ in the two copies of $\mathbb R$ in $X$. Decompose the mapping cylinder into the two open sets $A$ and $B$ which are the complements of the two "bad" points in $X$ (regarding $X$ as a subspace of the mapping cylinder). Taking a little care with the point-set topology, one can check that $A$, $B$ and $A\cap B$ each deformation retract onto the circle end of the mapping cylinder. Then van Kampen's theorem says that $\pi_1$ of the mapping cylinder, which is isomorphic to $\pi_1(X)$, is isomorphic to the free product of two copies of $\mathbb Z$ amalgamated into a single $\mathbb Z$.
An interesting fact about $X$ is that it is not homotopy equivalent to a CW complex, or in fact to any Hausdorff space. For if one had a homotopy equivalence $f:X \to Y$ with $Y$ Hausdorff then $f$ would send the two bad points of $X$ to the same point of $Y$ so $f$ would factor through the quotient space of $X$ obtained by identifying these two bad points. This quotient is just $\mathbb R$ and the quotient map $X \to \mathbb R$ is not injective on $\pi_1$, so the same is true for $f$ and $f$ can't be a homotopy equivalence.