[Math] Cellular Homology of the 3-Torus

algebraic-topology

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:

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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".

Best Answer

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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