Which path connected space has fundamental group isomorphic to the group of rationals? More generally, is every group the fundamental group of a space?
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6See also http://math.ucr.edu/home/baez/week286.html. – Cheerful Parsnip May 03 '11 at 20:00
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The title of this question is almost exactly the title of a paper by Shelah https://www.ams.org/journals/proc/1988-103-02/S0002-9939-1988-0943095-3/S0002-9939-1988-0943095-3.pdf – Dan Ramras Aug 22 '20 at 19:39
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See the Wikipedia page on Eilenberg-Mac Lane spaces for an even better statement: For every group $G$ there is a $CW$-complex $K(G,1)$ (unique up to homotopy equivalence) such that $\pi_1(K(G,1)) \cong G$ and $\pi_{n}(K(G,1)) = 0$ for all $n \neq 1$. This is also true for every other value of $1$ (to quote Mariano Suárez-Alvarez) and abelian $G$ and proofs of these statements can be found in almost all books on algebraic topology.
A nice and and rather explicit example for a space with fundamental group $\mathbb{Q}$ can be constructed using the theory of graphs of groups, see exercise 6 on page 96 of Hatcher's book.
In 1988, Shelah proved that there is no "nice" compact space with fundamental group $\mathbb{Q}$, where nice means metric, compact (hence separable) path connected and locally path connected. Indeed, Shelah has shown the fundamental group of a nice compact space is either finitely generated or has the cardinality of the continuum.
t.b.
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Worth pointing out perhaps these constructions give spaces which have discrete pi1. Their pi1 is isomorphic to Q as a group, not as a topological group. There may be a way to construct a non semilocally simply connected space having a desired topological group, or having Q, as its fundamental group. I think it becomes a delicate question, sensitive to your definition of fundamental group.See https://math.stackexchange.com/q/667035/16490, https://mathoverflow.net/q/54391/19860, https://mathoverflow.net/q/26680/19860, and https://arxiv.org/abs/math/0501482 for more discussion. – ziggurism Dec 04 '17 at 04:28
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Every group is the fundamental group of a space; a relatively easy choice of such a space is the presentation complex associated to a presentation. First, every group $G$ has a presentation with some generators $g_i$ indexed by some set $I$ and some relations $r_j$ indexed by some set $J$. Let $X$ be the wedge of $|I|$ circles; by Seifert-van Kampen we know that $\pi_1(X) \cong F_{|I|}$.
Now we will add some $2$-cells corresponding to the relations. First, note that every relation $r_j$ determines a homotopy class of paths in $X$, hence a subspace isomorphic to $S^1$ of $X$. Associated to such a subspace is an attaching map $b_j : S^1 \to B^2$, and we can form the adjunction space $X \cup_{b_j} B^2$ in order to attach the appropriate $2$-cell and kill the relation $r_j$ (proven by a second application of Seifert-van Kampen). Doing this for all relations gives the appropriate space.
Qiaochu Yuan
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2Nice! But, if the group is not finitely generated, is it so easy to give a presentation? For example, how is a presentation of rationals? – MBL May 03 '11 at 20:10
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9@MBL: given a group $G$ there is a canonical map $F_{|G|} \to G$ from the free group on the elements of $G$ to $G$. We can take the set of relations to be the kernel of this map. (I didn't say the presentation had to be nice!) – Qiaochu Yuan May 03 '11 at 20:15
