= Solution
For a <CW complex> $X$, let $X^k$ be its $k$-skeleton. The <cellular chain complex> is
$$
C_k^{\mathrm{cell}}(X)=H_k(X^k,X^{k-1};\mathbb Z)\cong\bigoplus_{\text{$k$-cells }e}\mathbb Z[e].
$$
The <isomorphism> uses <excision> and $H_k(D^k,S^{k-1})\cong\mathbb Z$, after choosing an <orientation> for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the composite
$$
H_k(X^k,X^{k-1})\xrightarrow{\partial}H_{k-1}(X^{k-1})
\longrightarrow H_{k-1}(X^{k-1},X^{k-2}).
$$
The <long exact sequences> of skeleton pairs imply $d_{k-1}d_k=0$. The <cellular boundary formula> describes its coefficient at a $(k-1)$-cell as the <mapping degree> of the attaching <sphere> after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The <cellular homology theorem> identifies the <homology> of this complex with <singular homology>: relative <homology> of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.
The filtration $\mathbb{CP}^0\subset\mathbb{CP}^1\subset\mathbb{CP}^2$ gives one cell in each of dimensions zero, two and four, since $\mathbb{CP}^k\setminus\mathbb{CP}^{k-1}\cong\mathbb C^k$. There are no odd-dimensional cells, so every cellular differential vanishes. Thus
$$
H_k(\mathbb{CP}^2;\mathbb Z)\cong\begin{cases}\mathbb Z,&k=0,2,4,\\0,&\text{otherwise}.\end{cases}
$$
The <complex projective line> has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its <cellular chain complex> again has zero differentials, giving
$$
H_k(\mathbb{CP}^1\times\mathbb{CP}^1;\mathbb Z)\cong\begin{cases}
\mathbb Z,&k=0,4,\\\mathbb Z^2,&k=2,\\0,&\text{otherwise}.
\end{cases}
$$
A <homeomorphism> induces <isomorphisms> on <homology groups>. The second groups have different ranks, and therefore
$$
\boxed{\mathbb{CP}^2\not\cong\mathbb{CP}^1\times\mathbb{CP}^1.}
$$
Now orient the <Complex projective plane> by its complex coordinates. The inclusion of its two-skeleton $\mathbb{CP}^1$ identifies the <fundamental class> of a projective line with a generator $h$ of $H_2(\mathbb{CP}^2;\mathbb Z)\cong\mathbb Z$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.
Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented <homology classes> are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their <tangent spaces> are the complex $w$-axis and complex $v$-axis, so the intersection is a <transverse intersection>. The ordered real bases
$$
(\partial_{\operatorname{Re}w},\partial_{\operatorname{Im}w}),
\qquad(\partial_{\operatorname{Re}v},\partial_{\operatorname{Im}v})
$$
concatenate to the complex <orientation> of the ambient four-manifold. Thus the local sign is $+1$, and
$$
\boxed{h\cdot h=[L_0]\cdot[L_1]=1.}
$$
This computes the <intersection form> using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now gives
$$
\lambda(ah,bh)=ab,\qquad [\lambda]_{h}=(1),\qquad\det[\lambda]=1.
$$
Equivalently, the map $H_2\to\operatorname{Hom}(H_2,\mathbb Z)$ sending $a$ to $\lambda(a,-)$ is an <isomorphism>. This is exactly a <unimodular intersection pairing>, and proves <projective lines generate a unimodular intersection form>. Reversing the ambient <orientation> would change the matrix to $(-1)$ and would still be unimodular.
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