My question is how to prove $[\mathbb{CP}^\infty, \mathbb{CP}^\infty] \cong \mathbb{Z}$, where $\mathbb{CP}^\infty$ is the infinite complex projective space. I know that $[\mathbb{CP}^\infty, \mathbb{CP}^\infty] \cong H^2 (\mathbb{CP}^\infty ; \mathbb Z)$. Probably the argument for $[\mathbb{CP}^\infty, \mathbb{CP}^\infty] \cong \mathbb{Z}$ is via Eilenberg-MacLane spaces, but I can't figure it out.
Thanks for your input!
We can equip $\mathbb{CP}^{\infty}$ with a CW complex structure consisting of only a single cell in every even dimension. We can then compute $H^2(\mathbb{CP}^{\infty}; \mathbb{Z})$ using cellular cohomology: the cochain groups are $C^{2n}(\mathbb{CP}^{\infty}) \cong \mathbb{Z}$ and $C^{2n+1}(\mathbb{CP}^{\infty}) = 0$, so the differential is always zero, and hence the cohomology of the cochain complex is itself, i.e. $H^{2n}(\mathbb{CP}^{\infty}; \mathbb{Z}) \cong \mathbb{Z}$ and $H^{2n+1}(\mathbb{CP}^{\infty}; \mathbb{Z}) = 0$.
If you're not comfortable with cellular cohomology, here's a more indirect proof. First note that for a CW complex $X$, $\pi_{k-1}(X) = \pi_{k-1}(X^{(k)})$ where $X^{(k)}$ is the $k$-skeleton of $X$. The two-skeleton of $\mathbb{CP}^{\infty}$ is $\mathbb{CP}^1 = S^2$, and the three-skeleton is also $S^2$. So $\pi_1(\mathbb{CP}^{\infty}) = \pi_1(S^2) = 0$ and $\pi_2(\mathbb{CP}^{\infty}) = \pi_2(S^2) = \mathbb{Z}$. By Hurewicz's theorem, $H_1(\mathbb{CP}^{\infty}; \mathbb{Z}) = 0$ and $H_2(\mathbb{CP}^{\infty}; \mathbb{Z}) = \mathbb{Z}$. Finally, by the Universal Coefficient Theorem,
$$H^2(\mathbb{CP}^{\infty}; \mathbb{Z}) \cong \operatorname{Hom}(H_2(\mathbb{CP}^{\infty}; \mathbb{Z}), \mathbb{Z})\oplus\operatorname{Ext}(H_1(\mathbb{CP}^{\infty}; \mathbb{Z}), \mathbb{Z}) = \operatorname{Hom}(\mathbb{Z}, \mathbb{Z})\oplus\operatorname{Ext}(0, \mathbb{Z}) = \mathbb{Z}.$$