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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?

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This looks like a homework problem (at least, it was homework when I took a topology class). –  S. Carnahan May 21 '10 at 12:43
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In any case, if you are interested I have opened a thread on meta. –  Andrea Ferretti May 23 '10 at 12:23
    
I have edited the first lines, according to your permission. I hope they sound good to you. I have also removed the relevant comments. –  Andrea Ferretti May 23 '10 at 13:56
    
@Akela: I have removed the actual dialogue from meta. I cannot remove the whole thread, but I think now it should be better. –  Andrea Ferretti May 23 '10 at 18:15
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@Andrea Ferretti: Could you also please change the title? To something else, like "perceived differences", or "personal differences" or something like that? My idea of you as a person(outside this incident) is that you are a very gentle person. It would keep raking my conscience if it remains in writing that I accused you otherwise. I have deleted all the other comments to you except this one, so that this alone might remain. –  Akela May 24 '10 at 6:36
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6 Answers 6

up vote 22 down vote accepted

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.

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There is a projection map from $R^2 \setminus (0,0)$ down to this doubled line that simply forgets the y-coordinate except at x = 0. At x = 0 it projects y > 0 to the top origin and y < 0 to the bottom origin.

Using the open cover of the doubled line by two copies of R, one can show that this projection map is a fiber bundle with fiber R. Explicitly, there is a homeomorphism $$ f(x,y) = (x, y - x^2/4y) $$ from $R \times (0,\infty)$ to $R^2 \setminus \{(0,y): y \leq 0\}$ with inverse $$ g(x,u) = (x,(u + \sqrt{u^2 + x^2})/2). $$ (This may look more complicated than it really is; the function f takes the lines $y = c$ and turns them into a sequence of parabolae foliating $R^2 \setminus \{(0,y): y \leq 0\}$.

Thus the map from R2 minus the origin to the doubled line is a weak homotopy equivalence, and so the homotopy groups of the doubled line coincide with those of S1. The fundamental group is ℤ.

Some entertaining generalizations of this include the fact that any finite CW-complex accepts a weak homotopy equivalence to a space with only finitely many points.

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Is this a fibration or only a quasi-fibtration? (a quasi-fibration is still good enough for your argument). –  Chris Schommer-Pries May 21 '10 at 12:32
    
Weird. Is the universal cover just the usual real line? –  Pete L. Clark May 21 '10 at 12:39
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@Chris: The "usual" coordinate charts make it look like a quasifibration. I've added some details about a change of coordinates that makes it look more like an actual fibration. –  Tyler Lawson May 21 '10 at 12:43
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@Pete: No, it is not Hausdorff. You glue a bunch of real lines along open rays. –  S. Carnahan May 21 '10 at 12:45
    
@Scott: I can't quite see it yet. Could you be a little more more explicit? –  Pete L. Clark May 21 '10 at 12:58
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You can compute that the fundamental group is Z by considering the fundamental groupoid. The main reason that Van Kampen's theorem isn't applicable in this case is due to connectivity. This can be circumvented by considering the fundamental groupoid. This satisfies a push-out property, like the Van Kampen's theorem and using the standard open cover by two copies of the real line we see that the fundamental group is Z.

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Thanks! Very nice, once I figure it out fully. Could you also please provide a reference wherein such a Van-Kampen theorem for groupoids is considered –  Akela May 21 '10 at 12:56
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You can find it here: mathonline.andreaferretti.it/books/view/18/… –  Andrea Ferretti May 21 '10 at 14:03
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I recommend: Brown, R. "Groupoids and van Kampen's theorem." Proc. London Math. Soc. (3) 17 1967 385--401. This is one of the earliest references. –  Chris Schommer-Pries May 21 '10 at 15:45
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This one is short and crisp: math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf –  Peter Arndt May 21 '10 at 15:57
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There is also the book "Topology and Groupoids" by Ronald Brown. –  Kevin H. Lin May 22 '10 at 15:20
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Am I missing something ? I had a quick look at Massey and it does not seem to require any Hausdorff assumption for the Van Kampen theorem.

A simple way to compute the fundamental group of your "double line" is to build its universal cover, just as in the case of the circle.

Just take $R\times Z$ and make the following identifications:

For all $i\in Z$ and $x>0$, $(x,2i) \sim (x,2i+1)$

For all $i\in Z$ and $x<0$, $(x,2i) \sim (x,2i-1)$

Drawing a picture may help. Let us describe the covering map. It is the standard projection for all $(x,i)$ such that $x\neq 0$. The points $(0,2i)$ are sent to the upper origin and the points $(0,2i+1)$ are sent to the lower origin. So the fundamental group is isomorphic to $Z$ with generator $(x,i)\rightarrow (x,i+2)$.

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I think the issue is not being Hausdorff, but that the "standard" version of the Seifert-Van Kampen theorem requires connected intersection. –  Tyler Lawson May 21 '10 at 17:36
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May's reference is nice but doesn't explicitly compute the fundamental group from the information that the fundamental groupoid is a push-out. The result of the computation can be found in a paper of Andre Gramain, Le theoreme de van Kampen, http://www.numdam.org/item?id=CTGDC_1992__33_3_237_0 Apperently, this article is inspired from a forthcoming work of Bourbaki. :-)

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I don't quite see how to complete this argument, but here's an idea. Say your loop goes from 1 to $-1$ "along the top" and from $-1$ to 1 "along the bottom". A putative homotopy to the trivial loop is a continuous map from the closed square $S$ to your space $X$ with the property that 3 sides are mapped to 1 and the remaining side has two points mapped to your upper and lower origins, enclosing an interval mapped to the negative reals with the rest of the side mapped to the positive reals.

Then $S$ partitions into 4 disjoint sets: $U_+$ and $U_-$ the open preimages of the positive and negative numbers and $K_1$ and $K_2$ the closed (and so compact) preimages of the two origins. As $K_1$ and $K_2$ are compact, then they are a positive distance $\delta$ apart. It should be possible to find a path inside $S$ from a point in $U_-$ on the boundary of $S$ to a point on the opposite edge of $S$. This would give a contradiction.

To construct the path, split up $S$ into little squares of sidelength $<\delta/10$ say. None of these squares can hit both $K_1$ and $K_2$ and there should be some combinatorial way to take a chain of little squares missing both $K_1$ and $K_2$ between the two given points (here is where I'm waving my hands :-)).

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