Cellular homology of 3 Torus (Clarification) , Hatcher

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?

• Which part of this sentence is unclear to you? – 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.

• @CL sorry, the previous version wasn't quite right... – Kenny Wong Nov 13 '18 at 1:04
• 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. – CL. Nov 13 '18 at 8:19
• @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. – Kenny Wong Nov 13 '18 at 10:18