# CW complex structure of $\mathbb{CP}^1$

I'm trying to prove that $$\mathbb{CP}^1$$ has a CW complex structure.

The problem is that I'm confused about the formal definition of a CW complex (although Hatcher's intuitive explanations make perfect sense).

I know what the structure is supposed to be: $$\mathbb{CP}^1$$ is diffeomorphic to $$\mathbb{S}^2$$, which is the attachment of a $$2$$-cell to a point, so I should think of attaching $$U_1:=\{(z_0:z_1)\in\mathbb{CP}^1\mid z_1\neq 0\}$$ (i.e., $$\mathbb{S}^2$$ minus the north pole) to the point at infinity $$(1:0)$$ (north pole).

I've constructed the map $$\phi:\mathbb{D}^2\to \mathbb{CP}^1$$ with $$z\mapsto (z:\sqrt{1-|z|^2})$$, which maps the interior of $$\mathbb{D}^2$$ to $$U_1$$ and the boundary $$\partial\mathbb{D}^2$$ to $$(1:0)$$.

This looks exactly like what I need. But is this enough? If not, what else do I need to prove?

Well if you know that $$S^2\cong \mathbb C P^1$$ you only need a CW-structure on $$S^2$$.
For that you need a homeomorphism $$D^2/S^1\to S^2$$ and a proof that the following square is a pushout (where $$N\in S^2$$ is, say, the north pole) :
$$\require{AMScd} \begin{CD} S^1 @>>> \{N\} \\ @VVV @VVV\\ D^2 @>>> S^2 \end{CD}$$
The proof that it is a pushout is the same as the one that $$D^2/S^1$$ is homeomorphic to $$S^2$$.
(note that since everything in sight is compact, and $$S^2$$ is Hausdorff this makes things easier : you just need to define a continuous bijection $$D^2/S^1\to S^2$$)