Jtdylktuy

This is on pg 143 and is also asked here which I quote:

We given $T^3$ the $3$-torus a cell decomposition as follows:

$1$ $3$-cell , $3$ $2$-cell, $3$ $1$-cell and $1$ $0$-cell. Giving cellular chain
$0 to mathbb{Z} overset{d_3} longrightarrow mathbb{Z}^3 overset{d_2}longrightarrow mathbb{Z}^3 overset{d_1}longrightarrow mathbb{Z} to 0$

My question is also computing $d_3$. Hatcher showed previously it suffices to compute the degree of attaching map. What I do not understand, in particularly, this line on pg 143,

Each $Delta_{alpha beta}$ maps the interiors of two opposite faces othe cube homeomoprhically onto the complement of a point in the target $S^2$ and sends the remanining four faces to this point.

May someone elaborate on explicitly what this means?

[Corrected: previous version contained error in description of the attaching map.]

$Delta_{alpha beta}$ is the map $partial D_alpha^3 to S_beta^2$ obtained by composing the following two maps:

• the attaching map $partial D_alpha^3 to X^2$ from the boundary of the $alpha$th 3-cell to the 2-skeleton of the whole space;




• the quotient map $X^2 to S^2_beta$ collapsing the complement of the $beta$th 2-cell to a point.

(See page 141 of Hatcher.)

This is exactly what is described in the sentence that you highlighted.

• The cube is the only 3-cell. Its boundary is the surface of the cube. The 2-skeleton of the space is also the surface of the cube, except that opposite faces are identified. The attaching map from the boundary of the 3-cell to the 2-skeleton is projection map on the surface of the cube that identifies opposite faces.




• There are three 2-cells, indexed by $beta$. A 2-cell is represented in your diagram as a pair of opposite faces on the cube, which are identified with each other. The complement of this 2-cell consists of the other two pairs of faces. So the quotient map on the 2-skeleton collapses these other two pairs of faces to a point, while preserving the pair of faces that make up our chosen 2-cell. What remains of the 2-skeleton after this collapse is a 2-sphere. (Remember, the two faces in our chosen 2-cell which survive this collapse are really a single face, because they are identified; this is why we get a single 2-sphere after this collapse rather than a wedge product of two 2-spheres.) The interior of our chosen 2-cell maps onto the 2-sphere minus a single point, and this single point is the image of the other two pairs of faces.

So composing these two maps, we see that $Delta_{alpha beta}$ is a map from the surface of the cube to an $S^2$. There exists a point $p in S^2$ such that $Delta_{alpha beta}$ maps the interiors of each of the two faces making up our 2-cell homeomorphically to $S^2 setminus p $, and such that $Delta_{alpha beta}$ maps the other four faces to $p$. Viewing the surface of the cube as an $S^2$ too, the map $Delta_{alpha beta}$ can thought of as a map $S^2 to S^2$.

Added on request: How to explain that $Delta_{alphabeta} : partial D^3 cong S^2 to S^2$ has degree zero.

Fix a generator $[sigma] in H^2 (partial D^3) cong mathbb Z.$
To show that $Delta_{alpha beta}$ has degree zero, we must show that $(Delta_{alpha beta})_star ([sigma])$ is the zero element
in $H^2 (S^2)$. Let $r$ be the reflection $partial D^3 to partial D^3$ that exchanges the two faces making up the $beta$th 2-cell.
From our description of $Delta_{alpha beta}$ (the sentence highlighted in yellow in your original question), it is clear that
$Delta_{alpha beta} circ r = Delta_{alpha beta}$. This implies that
$ (Delta_{alpha beta})_star (r_star ([sigma])) = (Delta_{alpha beta})_star ([sigma]).$
But $r_star([sigma]) = - [sigma]$ since $r$ is a reflection on a sphere. So we have $- (Delta_{alpha beta})_star ([sigma]) = (Delta_{alpha beta})_star ([sigma])$,
hence $(Delta_{alpha beta})_star ([sigma]) = 0$. Therefore, $Delta_{alpha beta}$ has degree zero.