# CW complex.

#### skamoni

Let $$\displaystyle X$$ be the topological space obtained as the quotient of the sphere $$\displaystyle S^2$$ under the equivalence relation $$\displaystyle x \sim -x$$ for $$\displaystyle x$$ in the equatorial circle.

Describe a CW complex whose underlying space is $$\displaystyle X$$, and compute $$\displaystyle H(X)$$.

I know how to compute simplicial homology, I just can't describe the space in an adequate way, can anyone help me?

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#### TheArtofSymmetry

Let $$\displaystyle X$$ be the topological space obtained as the quotient of the sphere $$\displaystyle S^2$$ under the equivalence relation $$\displaystyle x \sim -x$$ for $$\displaystyle x$$ in the equatorial circle.

Describe a CW complex whose underlying space is $$\displaystyle X$$, and compute $$\displaystyle H(X)$$.

I know how to compute simplicial homology, I just can't describe the space in an adequate way, can anyone help me?
The space X is the real projective plane, $$\displaystyle \mathbb{R}P^2$$, which is the quotient space $$\displaystyle S^2/(x \sim -x)$$, where -x is the antipodal point of x for each $$\displaystyle x \in S^2$$.
The CW structure of $$\displaystyle \mathbb{R}P^2$$ has one cell $$\displaystyle e^k$$ in each dimension for k = 0, 1, and 2 such that $$\displaystyle e^0 \cup e^1 \cup e^2$$.
The attaching map for $$\displaystyle e^k$$ is the 2-sheeted covering projection $$\displaystyle \phi:S^{k-1} \rightarrow \mathbb{R}P^{k-1}$$ for k=1,2, where $$\displaystyle \mathbb{R}P^0$$ is a point and $$\displaystyle \mathbb{R}P^1$$ is a circle.

Computing a boundary map is not trivial. A good reference for this is Hatcher's "Algebraic Topology" p 137-148, especially Example 2.42.

The resulting cellular chain complex is as follows:

$$\displaystyle \cdots \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{2} \mathbb{Z}\xrightarrow{0} \mathbb{Z} \rightarrow 0$$

Now the homology group of $$\displaystyle X = \mathbb{R}P^2$$ is

$$\displaystyle H_0(X) = \mathbb{Z} \text{ }, H_1(X) = \frac{\mathbb{Z}}{2\mathbb{Z}},$$ and $$\displaystyle H_k(X) = 0$$ for $$\displaystyle k \geq 2$$.

#### skamoni

Hi thanks for your reply. Are you sure that the space is the real projective plane? I thought the real projective plane was constructed by identifying all antipodal points on $$\displaystyle S^2$$, whereas the space I'm looking at only identifies antipodal points on the equitorial circle.

#### TheArtofSymmetry

Hi thanks for your reply. Are you sure that the space is the real projective plane? I thought the real projective plane was constructed by identifying all antipodal points on $$\displaystyle S^2$$, whereas the space I'm looking at only identifies antipodal points on the equitorial circle.
If you only identifies antipodal points on the single equatorial circle, then it is a totally different question. I'll try to put up the solution once it is ready.

#### TheArtofSymmetry

Let X be the space you referred.

Then X has the CW structure with a single 0-cell, one 1-cell, and two 2-cells.

The projective plane I referred is homeomorphic with the space obtained from closed ball $$\displaystyle B^2$$ by identifying x with -x for each $$\displaystyle x \in S^1$$. So we needed only one 2-cell.

I think the space you referred requires two 2-cells attaching the boundary of each 2-cell to $$\displaystyle \mathbb{R}P^1$$.

The exact sequence looks like

$$\displaystyle 0 \rightarrow \mathbb{Z} \times \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \rightarrow 0$$.

Now the resulting homology group is

$$\displaystyle H_0(X)=\mathbb{Z}$$ and $$\displaystyle H_k(X)=0$$ for $$\displaystyle k \geq 1$$. Note that $$\displaystyle \mathbb{Z} \times \mathbb{Z} \xrightarrow{2} \mathbb{Z}$$ in the above exact sequence is surjective.

• skamoni