Let $M$ be a closed oriented smooth 4-manifold. Denote by $[M, \mathbb{C}P^{\infty}]$ homotopy classes of continuous maps from $M$ to $\mathbb{C}P^{\infty}$.
I would like to know how to show $$ [M, \mathbb{C}P^{\infty}]=[M,\mathbb{C}P^2].$$
I know that $\mathbb{C}P^2$ has (real) dimension 4 so the above equation looks plausible. But I would like to know a rigorous proof.
I am not familiar the complex projective space except its definition nor homotopy theory.
Thank you.
First, note that $\Bbb{CP}^2$ is a subcomplex of $\Bbb{CP}^\infty$, the latter having a CW structure with a single cell in every even dimension. Because smooth manifolds are triangulable, they have a CW structure, so cellular approximation says that (because $M$ has dimension at most $4$ - so all the cells in the triangulation are dimension at most $4$) every map $f: M \to \Bbb{CP}^\infty$ is homotopic to a map $g_1: M \to \Bbb{CP}^2$. If $f$ is also homotopic to $g_2: M \to \Bbb{CP}^2$, then $g_1$ and $g_2$ are also homotopic (again invoke cellular approximation - there are no cells in odd dimensions in our cell structure on $\Bbb{CP}^\infty$). So we have a canonical map $[M,\Bbb{CP}^\infty] \to [M,\Bbb{CP}^2]$; surjectivity is obvious and if two maps $f_i: M \to \Bbb{CP}^\infty$ are homotopic to the same map $M \to \Bbb{CP}^2$, they're homotopic, so the map is a bijection as desired.