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Are there two path connected topological spaces $X,Y$ such that the fundamental groupoid of $X$ is not isomorphic to the fundamental groupoid of $Y$ but the fundamental group of $X$ is isomorphic to the fundamental group of $Y$ ?

I guess that there exists such a pair of topological spaces. I don't know an example though. I am very interested to see such a pair.

Edit: The first version of the question was already solved by Zev Chonoles. Here is the second version of the question

Are there two path-connected topological spaces $X,Y$ such that:

1) $|X|=|Y|$

2) The fundamental group of $X$ is isomorphic to the fundamental group of $Y$

3) The fundamental groupoid of $X$ is not isomorphic to the fundamental groupoid of $Y$

In other words, this is a comparison between the fundamental groupoid functor and the combined use of the fundamental group and the forgetful functor from Top to Set

Thank you

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up vote 7 down vote accepted

It suffices to prove the following:

If $X$ and $Y$ are connected groupoids, then given any bijection $F : \operatorname{ob} X \to \operatorname{ob} Y$ and any group isomorphism $\Phi : X(x, x) \to Y(y, y)$ (where $x$ is in $X$ and $y = F x$), there is a groupoid isomorphism $X \to Y$ extending the given data.

Choose an isomorphism $p_{x'} : x \to x'$ for each $x'$ in $X$, with $p_x = \mathrm{id}_x$, and choose an isomorphism $q_{y'} : y \to y'$ for each $y'$ in $Y$, with $q_y = \mathrm{id}_y$. We define a functor $F : X \to Y$ as follows: $F$ acts on objects as the given bijection $\operatorname{ob} X \to \operatorname{ob} Y$, and for each morphism $f : x' \to x''$ in $X$, we define $$F f = q_{y''} \circ \Phi (p_{x''}^{-1} \circ f \circ p_{x'}) \circ q_{y'}^{-1}$$ It is easily shown that $F$ is indeed a functor and has the desired properties.

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I don't think the $p_{x'}$ and $q_{y'}$ should be called isomorphisms. And also, you don't even need the arrows $p_x$ and $q_y$ to be the identity. – mercio Jul 15 '13 at 9:52
$X$ and $Y$ are groupoids, so every arrow is an isomorphism – and I later need to invert them, so I may as well draw attention to the fact. If $p_x$ and $q_y$ are not the identity then $F$ does not extend $\Phi$ in general. – Zhen Lin Jul 15 '13 at 9:54
the 2 kinds of isomorphisms confused me briefly. And right I missed that you wanted to extend $\Phi$ (you still get a functor if you don't). – mercio Jul 15 '13 at 9:59
@Zhenlin Thanks for the answer. I will check the details later as I am busy now. However your answer is very surprising for me, as I was expecting a counterexample. – Amr Jul 15 '13 at 12:56
It shouldn't be surprising. As demonstrated above, a connected groupoid is completely characterised by the group of automorphisms at an object and the cardinality of its object set. There is no higher structure present – unlike a topological space! – Zhen Lin Jul 16 '13 at 0:48

One example would be with $X$ and $Y$ indiscrete spaces of cardinalities $1$ and $2$ respectively. Their fundamental groupoids have different cardinalities and therefore are not isomorphic, but both of their fundamental groups are trivial.

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+1 Thanks Zev. This answer suffices. I would be intrested to see other examples of spaces with the same cardinality )if any). – Amr Jun 14 '13 at 3:56
@Amr: I agree, I can't think of any examples off the top of my head where the cardinalities are the same. I'm looking forward to other answers that may provide such examples. – Zev Chonoles Jun 14 '13 at 4:01
Informally: I want to know whether the fundamental groupoid is better than the combined usage of the fundamental group and the forgetful functor from $Top$ to $Set$ – Amr Jun 14 '13 at 4:02
@Amr: I guess a slight generalization of the approach in my example is this: take a space $X$ where some finite number of points are naturally picked out as special in the fundamental groupoid (considered as a category), e.g. the initial objects, and then let $Y$ be the same space $X$ but with a point topologically indistinguishable from a special one added in, so that we can say the number of such special objects in $\Pi_1(X)$ and $\Pi_1(Y)$ are different and therefore they cannot be isomorphic. – Zev Chonoles Jun 14 '13 at 4:42
However, I don't think that points are going to be categorically-distinguishable in any way in the fundamental groupoid of a path-connected space, so that the only way there will be finitely many special points is if the space itself is finite, and therefore, this will again produce an example where $X$ and $Y$ have different cardinalities. – Zev Chonoles Jun 14 '13 at 4:43

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