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Let $G=\mathbf Z\times \mathbf Z$ and let $\langle x, y| \ xyx^{-1}y^{-1} \rangle$ be a presentation for $G$.

In Example 1.46 of Hatcher's Algebraic Topology, the author mentions that the presentation complex of $G$ is the torus $S^1\times S^1$.

I am not able to see why this is so.

The presentation complex here is formed by attaching the boundary of a $2$-disc on the wedge sum of two copies of $S^1$. The attaching map $\varphi:S^1\to S^1\vee S^1$ is determined by the relation $xyx^{-1}y^{-1}$. I am not even able to imagine the final picture we get after performing this attaching.

Can somebody please help me formally prove that the presentation complex at hand is actually the torus. Any geometrical insights are welcome.

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  • $\begingroup$ do you know what is $K(G,1)$ space??Ok I misread your qes...here you don't need $K(G,1)$ spaces...but just try to draw the polygonal representation corresponding to the group...and see what space you get from there $\endgroup$ – Anubhav Mukherjee May 26 '16 at 15:41
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Are you familiar with this picture?

Square

(courtesy of Wikimedia)

To get a torus you start with a full square $[0,1]^2$ and then you identify opposite edges in the way depicted above. Indeed if you first identify the red edges you get a cylinder, and then the blue edges become the top and the bottom circle. If $x$ is for blue and $y$ is for red, the corresponding word in the free group is indeed $xyx^{-1}y^{-1}$.

But if you start with the boundary of the square $\partial [0,1]^2$ instead of the full square, and you do the same identification, you get a wedge sum of two circles: they both go around the torus, one around the "small" radius and one around the "long" radius if you will.

To get back the full torus, you have to attach a full square to the boundary $\partial [0,1]^2$, and then identify opposite edges. This was the description in the first paragraph. But this is exactly the same thing as doing it in the reverse order: first identify opposite edges, and then glue a full square to the resulting wedge of two circles. Now we get the description that you are asking about: a wedge of two circles, and attach a $2$-cell (there's no real difference between a disk and a square here) along the word $xyx^{-1}y^{-1}$.


Here is another possibility: you can "thicken up" the wedge sum $S^1 \vee S^1$ to get this space (sorry for the crappily hand-drawn mess, if I have time I will do better):

The big clump to the left can be contracted to a point, and both "handles" can be contracted to a circle, so this deformation retract onto $S^1 \vee S^1$.

As you can see the boundary of this space is a circle. If you attach a disk along this circle, you get a torus, which is a bit easier to picture; and the corresponding word in $\pi_1(S^1 \vee S^1)$ is also $xyx^{-1}y^{-1}$ (follow along, you will see that you go once around the first handle, then once around the second handle, once around the first handle in reverse direction, and finally once around the second handle in reverse direction). This is called a "handlebody decomposition" of the torus.

I hope you now have a clearer picture of the torus.

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    $\begingroup$ BTW, this is pretty much as formal as this kind of proof can get. If you try to write everything down analytically you will go crazy. $\endgroup$ – Najib Idrissi May 26 '16 at 15:57
  • $\begingroup$ Thank you so much for the superb explanation. $\endgroup$ – caffeinemachine May 26 '16 at 16:09
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    $\begingroup$ It's not all that bad to write the proof down more formally, using basic theorems about quotient maps, as long as one has the first picture drawn out. $\endgroup$ – Lee Mosher May 26 '16 at 16:19

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