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
    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
    Jun 8, 2021 at 10:43
  • $\begingroup$ @Carrondo Exactly, it is a degree 2 map in particular. $\endgroup$
    – Rundasice
    Jun 8, 2021 at 13:25
  • $\begingroup$ @Rundasice Indeed it is, thank you! $\endgroup$
    – Oscar
    Jun 8, 2021 at 16:53

1 Answer 1


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
    Jan 27, 2021 at 7:11

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