Attaching maps for the CW-decomposition of the 3-torus I want to calculate the homology of the $3$-torus via cellular homology.
I figured out a CW-decomposition of the $3$-torus: $1$ $0$-cell, $3$ $1$-cell, $3$ $2$-cell, $1$ $3$-cell. So the chain complex looks like:
$0 \to \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \to 0$
The first differential is $0$ since there is only one $0$-cell. The second differential is also $0$.
So my question is what is the third differential? I can't figure out the attaching map, so I can calculate the degree.
 A: You need an understanding of the attaching map of the 3-cell in order to compute the differential $\mathbb{Z} \mapsto \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$.
Think of $T^3$ as the quotient of the cube $[-1,+1]^3$ by identifying $(x,y,-1) \sim (x,y,+1)$, $(x,-1,z)\sim(x,+1,z)$, and $(-1,y,z) \sim (+1,y,z)$. 
Let $q : [-1,+1]^3 \to T^3$ be the quotient map. 
The $0$-skeleton of $T^3$ consists of one $0$-cell, namely the image of the eight corners of the cube. The $1$ skeleton of $T^3$ consists of three $1$-cells, namely the images of the three sets of four parallel sides of the cube. The $2$-skeleton consists of three $2$-cells, namely the images of the three sets of two parallel faces of the cube.
Let's name the 2-cells of $T^3$:


*

*$\sigma_x = q \bigl( \{\pm 1\} \times [-1,+1] \times [-1,+1] \bigr)$

*$\sigma_y = q \bigl( [-1,+1] \times \{\pm 1\} \times [-1,+1] \bigr)$

*$\sigma_z = q \bigl( [-1,+1] \times [-1,+1] \times \{\pm 1\}  \bigr)$


Let's orient these 2-cells. Using the right hand rule, orient each face of the cube by letting your thumb point in the positive direction of the axis normal to that face; let's refer to these as the "positive orientations" of the six faces of the cube. For example, the faces 
$$\sigma^-_z = [-1,+1] \times [-1,+1] \times - 1 \quad\text{and}\quad\sigma^+_z = [-1,+1] \times [-1,+1] \times + 1$$
 are positively oriented by letting your thumb point in the positive $z$-direction. These two positive orientations are compatible with each other under the gluing, therefore they descend to the same orientation on the quotient 2-cell $\sigma_z \subset T^3$.
Let's compute the differential. The unique 3-cell of $T^3$ is parameterized by the entire cube $[-1,+1]^3$. The boundary of this cube is given the "outward orientation", defined by using the right hand rule with your thumb pointing outward from the inside of the cube to the outside. Notice: the outward orientation on $\sigma^+_z$ agrees with the positive orientation on $\sigma^+_z$, whereas the outward orientation on $\sigma^-_z$ disagrees with the positive orientation on $\sigma^-_z$. It follows that the $z$-coefficient of the difreremtial $\mathbb{Z} \mapsto \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$ equals $-1+1=0$. The $x$ and $y$ coefficients are zero for similar reasons.
The differential $\mathbb{Z} \mapsto \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$ is therefore the zero map.
A: This answer is beyond the scope or level of the question, but it may be of interest to some to note that there is a notion which reflects the geometry of the $3$-cube and $3$-torus better then the usual chain complex. This uses the crossed complex $\Pi X_*$ defined for a filtered space $X_*$, and in particular for a CW-complex $X$ with its skeletal filtration $X_*$. Here  $\Pi X_*$ is defined using the fundamental groupoid $C_1=\pi_1(X_1,X_0)$ and the relative homotopy groups $$C_n= \pi_n(X_n,X_{n-1},x), n \geqslant 2, x \in X_0,$$ and has boundaries $\delta_n: C_n \to C_{n-1}$ and operations of $C_1$ on $C_n$. This idea for the single point case goes back to Blakers (1948) and to Whitehead's paper "Combinatorial Homotopy II". 
In the crossed complex $\Pi I^3_*$, let $c^3$ be the top dimensional cell and for 
$\alpha=-,+$ let $c^\alpha_i=\partial^\alpha_i c^3$ be the two faces in direction $i$ , the first for $\alpha = -$, the last for $\alpha = +$. 
The formula for the boundary is then as follows: 
$$\delta_3 c^3= 
- c^+_3 -
( c^-_2)^{u_{2}c} - c^-_1 + ( c^-_3)^{u_{3}c} + c^+_2 +
( c^-_1)^{u_{1}c} 
 $$ (where $u_i = \partial^+_1  \cdots \hat{\imath}
\cdots
 \partial^+_{3}$). 
Notice that if we identify $c^-_i=c^+_i$ we get a non zero formula for the $3$-torus. This formula gets nearer to modelling the geometry of the attaching map of the $3$-cell of the $3$-torus.   If we cancel out the operations, and make it abelian, we get $0$, as given by Lee. 
This is a specialisation  of a formula for dimension $n$ known as the "Cubical Homotopy Addition Lemma", given in the EMS Tract vol 15 partially titled Nonabelian Algebraic Topology. 
