# Homology of the unit ball $D^3$ with identification of boundary points by $180^\circ$ degree rotation around the vertical axis

This problem comes from Topology and Geometry of Bredon :

Let $$X$$ result from $$D^3$$ by identifying points on its boundary $$S^2$$ taken into one another by the $$180^\circ$$ rotation about the vertical axis. Give $$X$$ a CW-complex structure and compute its homology.

Here is my concern : For two different CW-complex structures, I get different homology groups.

1. The first one consists of one $$0$$-cell, $$x$$, one $$1$$-cell, two $$2$$-cells $$\sigma_1, \sigma_2$$, and one $$3$$-cell $$\rho$$. See the figure below: Then the $$2$$-cell are attached to the $$1$$-cell by the word $$a^{\pm 2}$$ depending on the orientation, which are degree $$2$$ maps, and the $$3$$ cell is attached to each $$2$$-cell by a degree $$2$$ map since there are two identified preimages which are orientation preserving. is this correct? This gives $$H_1(X) \cong \mathbb{Z_2}$$, $$H_2(X) = 0$$, and $$H_3(X) = 0$$.

1. The second one consists of two $$0$$-cells, $$x_0, x_1$$, one $$1$$-cell $$a$$, one $$2$$-cell $$\sigma$$, and one $$3$$-cell $$\rho$$. See the figure below : Is the $$2$$-cell $$\sigma$$ -which is the left part of the sphere in the figure above- attached by the word $$aa^{-1}$$, and the $$3$$-cell a degree $$2$$ map since two preimage differ by a rotation, which is orientation preserving? I get then $$H_1(X) = 0$$ which seems weird because this space looks like a lens space except that there is no reflection.

• I noticed that this space is the unreduced suspension of $\Bbb{R}P^2$. Jan 27, 2021 at 2:46
• I'm doing the same problem right now and I'd like to clarify one thing... What's your attaching map for the 3-cell? Are you considering $f:S^2 \to X^{(2)}$ which takes $p$ and the point which differs from $p$ by 180º degrees to the point $p \in X^{(2)} \cong S^2$ (assuming $p$ is always the one with non-negative $x,y$ coordinates)? Jun 8, 2021 at 10:43
• @Carrondo Exactly, it is a degree 2 map in particular. Jun 8, 2021 at 13:25
• @Rundasice Indeed it is, thank you! Jun 8, 2021 at 16:53

First, I think you need to convince yourself that $$X\approx S(\Bbb{R}P^2)$$, the unreduced suspension of $$\Bbb{R}P^2$$ defined as the quotient space of $$\Bbb{R}P^2\times I$$ by collapsing $$\Bbb{R}P^2\times\{0\}$$ to a point and $$\Bbb{R}P^2\times\{1\}$$ to another point.

If you take $$\Bbb{R}P^2\approx D^2/{(x\sim-x)},\forall x\in S^1$$, then you will see that $$\Bbb{R}P^2\times I$$ is just the solid cylindrical space. Identifying the two copies indicated above, it forms the spherical objects that you drew.

Suppowe we're working with the standard cell structure, i.e., $$\Bbb{R}P^2=e_1^0\cup e_1^1\cup e_1^2$$ and $$I=[0,1]=e_2^0\cup e_3^0\cup e_2^1$$, then we can deduce the cell structure on $$S(\Bbb{R}P^2)$$: \begin{align} \text{0-cells: }e_2^0,\text{ }e_3^0\\ \text{1-cell: }e_1^0\times e_2^1\\ \text{2-cell: }e_1^1\times e_2^1\\ \text{3-cell: }e_1^2\times e_2^1\\ \end{align} because any $$e_1^\alpha\times e_\beta^0$$ gets quotient out by the definition of unreduced suspension.

At this point, you might realize that the second cell structure in your post agrees with this one, which yields a cellular chain complex $$0\to\Bbb{Z}\overset{\partial_3}{\to}\Bbb{Z}\overset{\partial_2}{\to}\Bbb{Z}\overset{\partial_1}{\to}\Bbb{Z}\oplus\Bbb{Z}\to 0$$ By direct computation $$\operatorname{im}(\partial_1)=\langle e_3^0-e_2^0 \rangle=\langle x_1-x_0\rangle$$, which kills one copy of $$\Bbb{Z}$$ in the 0-th cellular chain group. $$\ker(\partial_1)=0$$, which implies the triviality of $$H_1$$. Next, $$\operatorname{im}(\partial_2)=0$$ as argued by you in the post. $$\partial_3$$ is a multiplication by $$2$$. Now, just apply the definition of cellular homology to get the answer.

The first "cell structure" is not even a cell structure, notice that points in $$\sigma_1$$ need to be identified according to the rotation about $$z$$-axis, so it's not a "nice" cell attached to the $$1$$-skeleton. So number $$1$$ is invalid.

There is a significant difference between $$X$$ and the lens space. A Mayer-Vietoris sequence argument shows that for a complex $$K$$ $$H_{k+1}(S(K))\cong \tilde{H}_k(K)$$ In particular, $$H_1(S(\Bbb{R}P^2))\cong\tilde{H}_0(\Bbb{R}P^2)\cong 0$$, so there is nothing weird.

• Thank you for pointing out the unreduced suspension! It definitively helps. Jan 27, 2021 at 7:11