2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Which part of this sentence is unclear to you? $\endgroup$ – Cheerful Parsnip Nov 12 '18 at 23:22
2
$\begingroup$

[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.

$\endgroup$
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.