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:

CW-complex structure

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 :

CW-complex structure

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.

  • 1
    $\begingroup$ I noticed that this space is the unreduced suspension of $\Bbb{R}P^2$. $\endgroup$
    – Kevin.S
    Commented Jan 27, 2021 at 2:46
  • $\begingroup$ 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)? $\endgroup$
    – Oscar
    Commented Jun 8, 2021 at 10:43
  • $\begingroup$ @Carrondo Exactly, it is a degree 2 map in particular. $\endgroup$
    – Rundasice
    Commented Jun 8, 2021 at 13:25
  • $\begingroup$ @Rundasice Indeed it is, thank you! $\endgroup$
    – Oscar
    Commented Jun 8, 2021 at 16:53

2 Answers 2


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.

  • $\begingroup$ Thank you for pointing out the unreduced suspension! It definitively helps. $\endgroup$
    – Rundasice
    Commented Jan 27, 2021 at 7:11

For future readers... There is a more computational way to do this problem which doesn't rely on observations of the above nature and generalizes quite readily to various quotient spaces. There are still a couple nontrivial steps, however I do perceive this solution as being a bit more instructive in how to proceed systematically with problems like it. In general, a good technique for finding the homology of quotient spaces like this is to construct a CW-complex whose quotient is the appropriate CW-complex. This can be achieved via starting with a good cell decomposition of $\mathbb{D}^3 \approx 2e^0 \cup 2 e^1 \cup 2 e^2 \cup e^3$ with the "obvious" attaching maps to make a solid 3-sphere. The reader is invited to form this cell structure and look at the boundary maps. Notice that under the quotient map we obtain a cell decomposition for our space $X$ of the form $2e^0 \cup e^1 \cup e^2 \cup e^3$. The three observations we need for figuring out the degrees of the attaching maps are the following:

  1. The attaching map $\phi$ for the boundary of the 3-sphere is the suspension of a degree 2 map for the circle, which consequently has degree 2. I 'believe' there is a description of this exact argument in Hatcher. This could also be seen by a local degree argument. Take any point $p \in \text{im}\phi$ which is not the north or south pole and the entire sphere as a neighborhood of $p$. Then, the preimage under the attaching map has 2 elements contained in disjoint open sets (two hemispheres of $\partial \mathbb{D}^3$). The map $f$ restricted to each of these open sets is homotopic to the identity, thus has the same local degree. The latter, alternative argument is similar to Bredon's example 7.7 in section 4.
  2. The 2-cell we are left with after the identification has a boundary comprised of the 1-cell traversed in both directions, hence has an attaching map which induces a trivial boundary map.
  3. The boundary map for the 1-cell has not changed after the quotienting procedure.

Putting all of these pieces together, we obtain the following cellular chain complex

$... \rightarrow 0\rightarrow \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0}\mathbb{Z}\xrightarrow{\begin{bmatrix}1 \\ -1\end{bmatrix}} \mathbb{Z}^2 \rightarrow 0 \rightarrow ...$

whose homology is easy to calculate.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .