Let $X$ be an $n-1$ connected space. Why does a map $X\rightarrow K(\pi_n(X),n)$ that induces an isomorphism on $\pi_n$ exist and what is this map?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
If $X$ is a CW-complex, then you can build a $K(\pi_n(X),n)$ by attaching cells (of dimension $n+2$ and above) to $X$ to kill off the higher homotopy groups. The map you are thinking of is then an inclusion of $X$ into $K(\pi_n(X),n)$ as a sub-complex.
As stated, the statement is false. For example, let $X$ be the Hawaiian earring. Then $X$ is 0-connected. But $X$ can't have a map to an Eilenberg-Maclane space that's an isomorphism on $\pi_1$, because as a compact space it would have to have compact image, which would necessarily have finitely-generated $\pi_1$ in a CW complex like an Eilenberg-Maclane space, and $X$ has infinitely generated fundamental group. The statement is true for if $X$ is a CW-complex, as wckronholm explains.
Let me assume $n>1$ and —so that your claim is true— that $X$ is a CW complex.
The Hurewicz theorem, in view of the hypothesis, gives us an isomorphism $$\phi:H_n(X,\mathbb Z)\to\pi_n(X).$$ On the other hand, in view of the hypothesis made on $X$, the universal coefficient theorem for cohomology gives us an isomorphism $$\alpha:H^n(X,\pi_n(X))\to \hom(H_n(X,\mathbb Z), \pi_n(X)) $$ Moreover, there is a canonical bijection $$\beta:[X,K(\pi_n(X), n)]\to H^n(X,\pi_n(X))$$ with $[X,K(\pi_n(X), n)]$ the set of homotopy classes of maps $X\to K(\pi_n(X),n)$.
The map you are looking for is any $f:X\to K(\pi_n(X),n)]$ whose homotopy class $[f]$ is such that $\alpha(\beta([f]))=\phi$.
(This works for any $K(\pi_n(X),n)$ that you pick)