TITLE: What can be said about the map $K(n)_\ast(X) \to K(n)_\ast(\tau_{\leq n}X)$ when $X$ is a finite complex?
QUESTION [5 upvotes]: Ravenel and Wilson showed that $K(\mathbb Z / p^j,q)$ is $K(n)$-acyclic for any $q \geq n+1$, and that $K(\mathbb Z, q)$ is $K(n)$-acyclic for $q \geq n+2$. It follows that $K(A,q)$ is $K(n)$-acyclic for $q \geq n+2$ when $A$ is finitely-generated.
From here, a Serre spectral sequence argument reveals that the map $\tau_{\leq m} X \to \tau_{\leq n+1} X$ (where $\tau$ is Postnikov truncation) is a $K(n)$-local equivalence for all $m \geq n+1$ when $X$ has finitely-generated homotopy groups. (For $\pi$-finite spaces, $\tau_{\leq m} X \to \tau_{\leq n} X$ is in fact a $K(n)$-local equivalence, as observed by Carmeli,Schlank, and Yanovski).
It's tempting to conclude that $X \to \tau_{\leq n+1} X$ is a $K(n)$-local equivalence for any $X$ with finitely-generated homotopy groups, but this can't possibly be true. If it were true, then in particular $X \to \tau_{\leq n+1} X$ would be an equivalence for all simply-connected finite spaces $X$. Then we could conclude $K(n)_\ast(S^2) = K(n)_{\ast+n+1}(\Sigma^{n+1} S^2) = K(n)_{\ast+n+1}(pt)$, which is false.
Indeed, according to Bauer, the convergence of the spectral sequence for the Postnikov tower of $X$ only holds when $X$ is $n$-truncated. This leads to my
Question: If $X$ is a space with infinitely many nontrivial homotopy groups, is there any meaningful relationship between $K(n)_\ast(X)$ and $K(n)_\ast(\tau_{\leq n} X)$ or $K(n)_\ast(\tau_{\leq n+1} X)$? (Beyond the mere existence of a map -- for all I know, this map is zero!) How about if $X$ is finite? Or perhaps, what if $X$ is $(n-1)$-connected?

REPLY [6 votes]: Let me make more concrete a comment I already wrote. In [Adv. Math. 201 (2006), 318-378], my example 2.22, illustrating a theorem just before it when $n=1$, says that, for any spectrum $Y$, there is a short exact sequence of Hopf algebras over $K(1)_*$ as follows:
$$ K(1)_*(\mathbb PY) \rightarrow K(1)_*(\Omega^{\infty}Y) \rightarrow K(1)_*(\tau_{\leq 2}\Omega^{\infty}Y).$$
Here $\mathbb PY$ is the free $E_\infty$--algebra generated by $Y$: the wedge of all the extended powers of $Y$.  
So if $X = \Omega^{\infty}Y$, then $K(1)_*(X) \rightarrow K(1)_*(\tau_{\leq 2}X)$ is onto.  One could wonder if one has an epimorphism for other $X$.  Okay, lets try something else: Bousfield gave a cute short argument that if $X$ is $E_*$--acyclic, so is $K(\pi_j(X), j)$ for any $j$. So if $X$ is $K(1)_*$--acyclic, so is $\tau_{\leq 2}X$, and thus our map is still an epi.
This is enough fun for one answer, so I'll leave things here.