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:

enter image description here

$1$ $3$-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?

  • $\begingroup$ Which part of this sentence is unclear to you? $\endgroup$ – Cheerful Parsnip Nov 12 '18 at 23:22

[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 $\Delta_{\alpha\beta} : \partial D^3 \cong S^2 \to S^2$ 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.

  • $\begingroup$ @CL sorry, the previous version wasn't quite right... $\endgroup$ – Kenny Wong Nov 13 '18 at 1:04
  • $\begingroup$ Thank you so much! That really helped. But I am still uncertain what exactly the "projection map" is - why is the local degree not $+2$ or $-2$? My understanding is this: Fix some $p \in S^2$. $\Delta_{\alpha \beta}^{-1}(p)$ are two points of opposite faces. Then the local map at one face is simply the identity, the other is a reflection along an axis, hence has degree $1$ and $-1$ respectively. I would like to see how you explain this part. $\endgroup$ – CL. Nov 13 '18 at 8:19
  • $\begingroup$ @CL. I agree with your explanation, though I am struggling to formalise it in a way that doesn't lead to a circular argument. I think it's easier to abandon local degrees entirely - see my latest edit. $\endgroup$ – Kenny Wong Nov 13 '18 at 10:18

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