I am reading a proof of the result that every element of the first homology a closed, oriented 3-manifold $M$ can be represented by a knot in $M$. This seems to be a pretty standard result, but I am having trouble understanding the following line:
Since dim $M$=3, any map $f: M \to \mathbb{CP}^\infty$ is homotopic to a smooth map $f_0:M \to \mathbb{CP}^1$.
For context, I am reading from p. 107 (Proposition 3.4.2) of Geiges' Introduction to Contact Topology. At first, I thought the statement above follows from the fact that $\mathbb{CP}^\infty = K(\mathbb{Z},2)$ and $f: M^n \to S^k$ is nullhomotopic when $k > n$; however, I must be missing something, because I am also confused by a later remark stating that
it is not true that $[M, \mathbb{CP}^\infty] = [M, \mathbb{CP}^1]$, since a map $F: M \times [0,1] \to \mathbb{CP}^\infty$ is not, in general, homotopic rel$(M \times \{0,1\})$ to a map into $\mathbb{CP}^1$. However, we do have $[M, \mathbb{CP}^\infty] = [M, \mathbb{CP}^2]$.
Would someone mind explaining why the homotopy claim is true, and why that reasoning might lead someone into the misapprehension the remark is fending off?
Let $X$ and $Y$ be CW complexes. A map $\varphi : X \to Y$ is called cellular if $\varphi(X^{(k)}) \subseteq Y^{(k)}$ for all $k$; here $X^{(k)}$ and $Y^{(k)}$ denote the $k$-skeletons of $X$ and $Y$ respectively.
Suppose now that $X$ is an $n$-dimensional CW complex. An immediate corollary of the above theorem is that every map $f : X \to Y$ is homotopic to a map $f' : X \to Y^{(n)}$. In particular, if $M$ is a three-dimensional manifold, then every map $M \to \mathbb{CP}^{\infty}$ is homotopic to a map $M \to \mathbb{CP}^1$ as $(\mathbb{CP}^{\infty})^{(3)} = \mathbb{CP}^1$.
Regarding the second remark, suppose $f : X \to Y$ and $g : X \to Y$ are homotopic. By the above theorem, there are maps $f' : X \to Y^{(n)}$ and $g' : X \to Y^{(n)}$ with $f$ homotopic to $f'$, and $g$ homotopic to $g'$. While $f'$ and $g'$ are homotopic as maps $X \to Y$, they are not necessarily homotopic as maps $X \to Y^{(n)}$; on the other hand, if two maps $X \to Y^{(n)}$ are homotopic, then they are homotopic as maps $X \to Y$ as $Y^{(n)} \subseteq Y$. For every $k$, there is a map $[X, Y^{(k)}] \to [X, Y]$, and for $k \geq n$ it is surjective by the Cellular Approximation Theorem, but it may not be injective: we could have $[f'] \neq [g']$, but $[f] = [g]$. As in this answer, given a homotopy $H : X\times[0,1]\to Y$ between $f$ and $g$, one can use the Cellular Approximation Theorem to produce a homotopy $H' : X\times[0,1] \to Y^{(n+1)}$ between $f'$ and $g'$ - note, $X\times[0,1]$ is a CW complex of dimension $n+1$. Therefore, the map $[X, Y^{(k)}] \to [X, Y]$ is injective for $k \geq n + 1$; in particular, identifying a map $X \to Y^{(n+1)}$ with the corresponding map $X \to Y$, we have $[X, Y] = [X, Y^{(n+1)}]$. Returning to your example, we have $[M, \mathbb{CP}^{\infty}] = [M, \mathbb{CP}^2]$ as $(\mathbb{CP}^{\infty})^{(4)} = \mathbb{CP}^2$.
Here's an example to keep in mind regarding the extra dimension. Consider $f : S^1 \to D^2$ and $g : S^1 \to D^2$ given by $f(z) = z$ and $g(z) = -z$. As $D^2$ is contractible, $f$ and $g$ are homotopic. On the other hand, $f$ and $g$ are already cellular, and regarded as maps $S^1 \to S^1$, they are not homotopic since they induce different maps on the fundamental group. Note that $(D^2)^{(1)} = S^1$ and $[S^1, D^2] \neq [S^1, S^1]$, but the natural map $[S^1, S^1] \to [S^1, D^2]$ is surjective as predicted.