It is not true that every map from $\mathbb CP^{\infty}$ to a finite complex is null-homotopic. For example it is a result due to Brayton-Gray that there are non-null maps $\mathbb CP^{\infty} \to S^3$, this story is told in P. May's "More Concise Algebraic Topology" (see cor. 2.4.3 on page 40). However, the examples constructed there are *phantom maps*, i.e. they become trivial when restricted to any finite subcomplex $\mathbb CP^k \subset \mathbb CP^{\infty}$. (A way to think about this is that it is just not possible to find a compatible choice of null-homotopies.) So my question is:

**Does there exist a finite CW-complex $X$ and a non-phantom map $\mathbb CP^{\infty} \to X$?**

Here are some thoughts:

(1) One can show that any such map has to be trivial on cohomology (with $\mathbb Z$ coefficients, using the cup product structure of $\mathbb CP^{\infty}$) and consequently also on homology (here we have to use that $X$ is finite). Lifting the map to the universal cover $\tilde{X}$ of $X$ and using Hurewicz, one can also see that the map is zero on all homotopy groups ($\mathbb CP^{\infty}$ only has one). However, there are plenty of non-trivial maps inducing zero on all homotopy and homology groups, the easiest example is maybe given by the composition $T^3 \to S^3 \to S^2$ where the first map collapes the complement of a ball and the second one is the Hopf map.

(2) Using the Sullivan conjecture/Miller's theorem one can probably show that the $p$-completion of any such map is null.

(3) Going in the opposite direction, maybe one can start with a non-trivial map $\mathbb CP^{k} \to X$ which induces zero on all homotopy and homology groups and then somehow prove that it extends to $\mathbb CP^{\infty}$. However I do not see how one could show in practice that extending the map is possible...