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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, 2013 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, 2013 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, 2013 at 3:15
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    $\begingroup$ What exactly are the attaching maps in the $T^3$ case? $\endgroup$
    – leeabarnett
    Apr 5, 2013 at 3:21
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    $\begingroup$ For the front face in your diagram, the corresponding $2$-cell attaches (starting at the bottom left corner) to $a$ following the arrow, to $c$ following the direction of the arrow, to $a$ opposite the direct of the arrow, and then to $c$ opposite the direct of the arrow. Then using the cellular boundary formula and quotienting out everything but $a$, the map goes across $a$ following the arrow and then reverses back to the original point. This is homotopic to the identity, so it has degree 0. Does this help? $\endgroup$
    – Potato
    Apr 5, 2013 at 3:30

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