CW-complex decomposition of $\mathbb{C}P^4/\mathbb{C}P^2$

Question

I want to calculate the cellular homology of $\mathbb{C}P^4/\mathbb{C}P^2$, but I have trouble finding a suitable CW-decomopistion.

What is a CW-complex decomposition of $\mathbb{C}P^4/\mathbb{C}P^2$?

Because $\mathbb{C}P^4 = S^5/S^1,\,\,\mathbb{C}P^2 = S^3/S^1,$ so $$ \mathbb{C}P^4/\mathbb{C}P^2 = S^5/S^3 \approx\ ?$$

Answer 1

From Hatcher's Algebraic Topology, page 8:

Quotients. If $(X, A)$ is a CW pair consisting of a cell complex $X$ and a subcomplex $A$, then the quotient space $X/A$ inherits a natural cell complex structure from $X$. The cells of $X/A$ are the cells of $X − A$ plus one new $0$-cell, the image of $A$ in $X/A$. For a cell $e^n_{\alpha}$ of $X - A$ attached by $\varphi_{\alpha} : S^{n-1} \to X^{n-1}$, the attaching map for the corresponding cell in $X/A$ is the composition $S^{n-1} \to X^{n-1} \to X^{n-1}/A^{n-1}$.

Note that $\mathbb{CP}^n$ is a CW complex with a single cell in every even dimension betweeen $0$ and $2n$ (inclusive). The attaching map of the $(2k+2)$-cell is the projection map for the $S^1$-fibre bundle $S^{2k+1} \to \mathbb{CP}^k$. In particular, for $m < n$, $\mathbb{CP}^m$ is a subcomplex of $\mathbb{CP}^n$ consisting of the cells in dimensions between $0$ and $2m$ (inclusive).

So $\mathbb{CP}^n/\mathbb{CP}^m$ is a CW complex with a single $0$-cell, and a single cell in every even dimension between $2m+2$ and $2n$ (inclusive).

In particular, $\mathbb{CP}^4/\mathbb{CP}^2$ is a CW complex with a $0$-cell, a $6$-cell, and an $8$-cell. The $6$-cell is attached to the $0$-cell via a constant map, so the six-skeleton is $S^6$. The seven-skeleton is also $S^6$ and the $8$-cell is attached to it via the map $S^7 \to \mathbb{CP}^3 \to \mathbb{CP}^3/\mathbb{CP}^2 = S^6$. It follows from the answers to this question that this map is not homotopic to a constant map.