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I'm trying to compute the homology of the 3-torus $T^3=S^1 \times S^1 \times S^1$. Trying to use the typical construction the 2-torus $T^2$ as a starting point, I identified pairs of opposite faces on a cube as shown below:

enter image description here

This gives a single 0-cell, three 1-cells, three 2-cells, and a single 3-cell. Still using my $T^2$ computations as a guide, I've gotten that the cellular chain complex for this space is

$0 \to \mathbb{Z} \overset{d_3} \longrightarrow \mathbb{Z}^3 \overset{d_2}\longrightarrow \mathbb{Z}^3 \overset{d_1}\longrightarrow \mathbb{Z} \to 0$

Since there is only one 0-cell, I get that $d_1=0$. I'm having trouble understanding how to compute $d_2$ and $d_3$. Could anyone shed some light on this?

For reference, this is problem IV.10.1 in Bredon's "Topology and Geometry".

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This is done as example 2.39 on page 142 of Hatcher's Algebraic Topology. This book is freely and legally available on the author's website, and you may find the relevant chapter here. (Warning: The link is to a large .pdf file.)

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  • $\begingroup$ I saw that example, but I'm having trouble understanding specifically why $d_2=0$; he refers to the construction in the $T^2$ calculation but I couldn't find where he did that. $\endgroup$ – leeabarnett Apr 5 '13 at 3:11
  • $\begingroup$ Use the cellular boundary formula. In the case of $T^2$, consider it as a square with opposite sides identified. Then the attaching map for the $2$-cell is $aba^{-1}b^{-1}$, so its boundary is the the kernel (the degree of $a$ and $a^{-1}$ cancel, etc) $\endgroup$ – Potato Apr 5 '13 at 3:15
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    $\begingroup$ Maybe that's bad notation. Doing the calculation for the 2-torus should make things clear. $\endgroup$ – Potato Apr 5 '13 at 3:15
  • $\begingroup$ Try that and let me know if you have difficulty. $\endgroup$ – Potato Apr 5 '13 at 3:17
  • $\begingroup$ What exactly are the attaching maps in the $T^3$ case? $\endgroup$ – leeabarnett Apr 5 '13 at 3:21

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