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 \(\displaystyle k \geq 2\).