Dear all, it seems any domain in Euclidean space, that is, connected open set will not make it. Am I right or could anyone give me a reference??
What should I do if I want to construct a space with fundamental group A5?
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Sign up to join this communityDear all, it seems any domain in Euclidean space, that is, connected open set will not make it. Am I right or could anyone give me a reference??
What should I do if I want to construct a space with fundamental group A5?
Your question has been answered, but I thought I'd add something. For any space $X$, $H_1(X,\mathbb{Z})$ is just the abelianization of $\pi_1(X,*)$. So, to find an example of a space with zero first homology but non-trivial fundamental group, you just need to find a space $X$ with $\pi_1$ equal to its commutator subgroup $[\pi_1,\pi_1]$ (such groups are called perfect groups). Then $H_1(X,\mathbb{Z})\simeq \pi_1^{ab} = \pi_1/[\pi_1,\pi_1] = 0$. These spaces are guaranteed to exist - see, for example, Gabriel's comment.
An example of such a space $X$, as has already been mentioned, is the homology 3-sphere. This is the quotient of $S^3$ (viewed as the group of unit quaternions) by the binary icosahedral group $2I$, a perfect (sub)group. In fact, this quotient map is a covering and exhibits $S^3$ as the universal cover of $X$. Hence $\pi_1\simeq 2I$ and $H_1 = 0$.
For any group $G$ there is a simple explicit construction of a 2-dimensional (connected) CW-complex $X$ such that $\Pi_1(X)=G$. This can be found in [Hatcher: Algebraic Topology, Corollary 1.28].
As an example take the group $G = A_5$ mentioned in the question. Then proceed the following steps:
For each relation attach a 2-cell with attaching map just the relation-loop, i.e. we attach
In summary we have $X=S^1_a \vee S^1_b \coprod_{a^2} D_1 \coprod_{b^3} D_2 \coprod_{(ab)^5} D_3$. It has one 0-cell, two 1-cells and three 2-cells.
In general each generator of $G$ corresponds to one copy of $S^1$ and each relation corresponds to one attached 2-cell.
If you want to look for such a space, look at the Wikipedia entry for Homology Sphere. The conjecture in the first sentence is false (take any homology sphere, embed it into some $\mathbb{R}^n,$ then thicken it to be an open set).