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I am reading Allen Hatcher's Algebraic Topology, and am trying to understand the proof to corollary 1.28:

For every group $G$ there is a $2$-dimensional cell complex $X_G$ with $\pi_1(X_G)\cong G$.

Since his book is available on his homepage, I will copy his proof below:

Choose a presentation $G=\langle g_\alpha\;|\;r_\beta\rangle$. This exists since every group is a quotient of a free group, so the $g_\alpha$s can be taken to be the generators of this free group with the $r_\beta$s generators of the kernel of the map from the free group to $G$. Now construct $X_G$ from $\bigvee_\alpha S_\alpha^1$ by attaching $2$-cells $e_\beta^2$ by the loops specified by the words $r_\beta$.

Which loops are specified here, and where do we attach the $2$-cells?

Edit: An example I thought of was a simple case where $G=\langle a, b\;|\;ab\rangle$, then we'd have $S_a^1\vee S_b^1$, and one $2$-cell which should "wrap around" both $S^1$? But then $\pi_1(X_G)$ would be trivial (contractible), unlike $G$?

Additionally, would there be an algorithm to work backwards, i.e. given a $2$-dimensional cell complex $X_G$ with some conditions (cells are disjoint except for the middle of the wedge), we can obtain $G=\langle g_\alpha\;|\;r_\beta\rangle$?

Szmagpie
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  • Every word in the generating set $g_\alpha$ determines an element of the fundamental group of the wedge of the circles. This gives you the loop (unique up to based homotopy). Now, attach a 2-cell along this loop (as in the definition of a CW complex, the word "where" does not really make sense). – Moishe Kohan Apr 18 '16 at 17:01

1 Answers1

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Given a group $G$, you could choose a set of generators $g_i$ and relators $r_i$ for the generators. Then construct a wedge of circles $X = \bigvee S^1$ with one circle for each generator $g_i$. The fundamental group of this wedge of circles is the free group $F(g_1, g_2, \cdots)$ on the generators of $G$.

Now $G$ can be realized as the quotient of $F(g_1, g_2, \cdots)$ by the subgroup $\langle r_1, r_2, \cdots \rangle$ generated by the relators. So if you attach a disk $D^2_i$ to $X$ by gluing the boundary $\partial D^2_i$ by the word $r_i$, for each word, then van Kampen's theorem would tell you that the fundamental group of the resulting 2-complex $X_G$ is precisely $G$. Basically we're pinching off the loops corresponding to $r_i$, so we get relators $r_i = 1$ added to the free group, which is precisely our group $G$.$^{(\bigstar)}$

For $\langle a, b | ab \rangle$, note that this is just isomorphic to $\Bbb Z$, because $ab = 1$ says $b = a^{-1}$. If you attach a disk to $S^1 \vee S^1$ along the loop $\partial D^2 \to S^1 \vee S^1$ given by the word $ab$, you're not going to get the trivial group: e.g., the loop going once around one of the circles in the wedge (i.e. corresponding to the word $a$) is nontrivial. In fact, it is going to be $\Bbb Z$, by van Kampen's theorem.

I don't understand your last question. If you have a cell complex, you can always compute it's fundamental group using van Kampen's theorem. It is precisely the fundamental group of the 1-skeleton (which is a wegde of circles upto homotopy equivalence) modulo the words the 2-disks are attached to the 1-skeleton by. This is all in Hatcher chapter 1.2 I believe.

$^{(\bigstar)}$Here's a nontrivial example which might help you understand the construction.

Take $G = \Bbb Z \oplus \Bbb Z = \langle a, b | aba^{-1}b^{-1} = 1\rangle$. Then we start with the wedge $S^1_a \vee S^1_b$ of two circles corresponding to $a$ and $b$ respectively. That has fundamental group $F(a, b) \cong \Bbb Z*\Bbb Z$ the free group on two generator. Now I want to glue a 2-disk along $aba^{-1}b^{-1}$. So start by taking a square (that's a 2-disk) with top and bottom labelled by $a$ equipped with an arrow (orientation) and the left and right sides labelled by $b$ equipped with some arrow (orientation). If you try to paste the square to the wedge of circles by $aba^{-1}b^{-1}$ you'll see you'll end up getting a torus. That indeed has fundamental group $G$. Visually, all we did was to make the loop corresponding to $aba^{-1}b^{-1}$ nullhomotopic, so in the fundamental group that gives me $aba^{-1}b^{-1} = 1$, i.e., $ab = ba$.

Balarka Sen
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  • Thanks for the answer! I have one question, if you are able to help. In the first example, when you add the $2$-cell, do you not "fill in" the circles and get $D^2\vee D^2$? When I visualize the attaching map, I see a graph with two "holes" become a shape without "holes" – Szmagpie Apr 18 '16 at 21:46
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    @Szmagpie No, it doesn't just fill in the circles. To do that you'd need to use $2$ disks. But here you are attaching just a single disk. The space you get by attaching $D^2$ to $S^1 \vee S^1$ along $ab$ is not so easy to visualize, actually. Maybe you could think of it as follows. Look at the figure 8, but think of it first as an S by deleting a bit of the self-intersection. Now glue in a triangle by gluing the base to the figure S in the obvious way. Then you make it self-intersect. So the resulting thing will look approximately like a cone on the figure 8 but self intersecting. – Balarka Sen Apr 18 '16 at 21:53
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    That you can see does not have fundamental group $0$. The loop corresponding to $a$ cannot be "pushed through the disk" (careful: it can be done so if you take the literal cone over 8, but not here because the disk is merely self intersecting near the 'x' point when you try to visualize it in R^3 - there's not really any intersection or anything). – Balarka Sen Apr 18 '16 at 21:55