# Can the fundamental group of any manifold be realized as the fund grp of a finite space?

Recently, I was asked to calculate the fundamental group of the space $X= \{a,b,c,d\}$ with open sets generated by $\{ a, c, abc, acd \}$.

Turns out, $\pi_1(X)\cong \mathbb Z$ and in fact, $X$ is the quotient of $S^1$ (with the northern and southern hemispheres identified). But the result was not so easy to prove and this motivates the questions:

• Is the fundamental group of every connected manifold (other restrictions / generalizations on the manifold are welcome) the fundamental group of a finite space? (Of course, it would not be Hausdorff). (I observe that there are many redundant points on a manifold where homotopy-equivalent loops need not traverse)

• Is calculating $\pi_1$ of such finite spaces easier than for the given space? (If yes, this gives a method to calculate fundamental groups of many familiar spaces)

Perhaps the fact that -- $\pi_1$ of any CW complex just depends on its 1-skeleton [edit:2 skeleton]-- may be helpful.

• You mean the 2-skeleton, right? – Ryan Reich Nov 10 '10 at 13:22
• I never paused to consider whether $\pi_1(X)$ could be nonzero for non-Hausdorff finite spaces. Is there a simple way of describing your space $X$ as a quotient of $S^1$? You mentioned identifying the two hemispheres in some way!? – Jim Conant Nov 10 '10 at 15:05
• It looks like he means that a is the northern hemisphere (collapsed to a single point), c is the southern hemisphere (likewise collapsed), and b and d are 1 and -1. Then the four sets he mentioned are the two open hemispheres and the complements of the two other points, which are indeed (with $a \cup b$ and $\emptyset$) the saturated open sets for the quotient of $S^1$ collapsing each hemisphere to different points. – Ryan Reich Nov 10 '10 at 15:58
• Sorry I had to be away. Ryan is right, the points on S^1 with positive (resp negative) Y-coordinate are all identified. The space is like, {1, -1, N, S}. @Ryan, I believe it is the 1-skeleton. A loop inside a disk (disk in R^2) can be homotoped to a loop on the boundary S^1. Am I committing some obvious blunder here? – Abhishek Parab Nov 10 '10 at 16:59
• Well, the fundamental group of a 1-skeleton is free. The 2-skeleton provides the relations, as you have demonstrated. – Ryan Reich Nov 10 '10 at 17:04