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Since $\Omega K(G, n+1)$ is a $K(G,n)$, we have a CW approximation/homotopy equivalence $K(G,n) \xrightarrow{\sim} \Omega K(G,n+1)$. The adjoint of this map is a map $\Sigma K(G,n) \to K(G,n+1)$. What is this map? Is it a homotopy equivalence?

Hatcher [SSAT, ch.2] asserts that "the Freudenthal suspension theorem implies that [this map] induces an isomorphism on homotopy groups up to dimension approximately $2n$", but I don't see how to apply the Freudenthal suspension theorem here.

I know that $\Sigma K(G,n)$ and $K(G, n+1)$ are homotopy equivalent, but I don't know why the adjoint map gives a homotopy equivalence [EDIT: As Andreas pointed out, it's not.]. (Or if it is at all: Hatcher seems to suggest that the map may fail to be an isomorphism for higher dimensional homotopy groups, which is odd since the two spaces has trivial homotopy in higher dimensions. Curiously, we can apply the Freudenthal suspension theorem for the homomorphism $\pi_i(K(G,n)) \to \pi_{i+1}(\Sigma K(G,n))$, and this is an isomorphic for dimensions $< 2n-1$.)

All this is supposed to be an easy observation of the homotopy groups of the Eilenberg-Maclane spectrum, so I think I'm missing something obvious. On that note, why do we not simply define the map $\Sigma K(G,n) \to K(G,n+1)$ to be some homotopy equivalence between the two $K(G,n+1)$'s [EDIT: This is completely wrong. Whoops.], but instead define it to be the adjoint of a homotopy equivalence?

EDIT: So I made a really silly mistake in thinking that $\Sigma K(G,n)$ is a $K(G, n+1)$, as pointed out in the comments. I'm keeping the offending sections of the question, so the comments make sense, but I would still like to know why the map $\Sigma K(G,n) \to K(G,n+1)$ induces isomorphisms in homotopy in low ranges.

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    $\begingroup$ Life would be so simple if $\Sigma K(G,n)$ were homotopy equivalent to $K(G,n+1)$. Since the circle is a $K(\mathbb Z,1)$, it would follow that the $n$-sphere is a $K(\mathbb Z,n)$, and then even I could compute the homotopy groups of spheres. $\endgroup$ – Andreas Blass Sep 20 '14 at 23:43
  • $\begingroup$ @AndreasBlass I knew I was forgetting something! I guess I'm back to where I started: if only I knew something more about the adjoint map... $\endgroup$ – JHF Sep 21 '14 at 0:52
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I think I managed to figure it out. I'm not sure if this answer interests anybody else, but I'm including it below for completeness (and possibly to solicit pointers on where my reasoning is going astray, again...). Please feel free to post your own answers or even comments on this answer to elucidate parts of the argument where I'm too hasty.

Fix a homotopy equivalence $f: K(G,n) \xrightarrow{\sim} \Omega K(G,n+1)$. Consider the composite of the maps: \begin{equation*} [S^i, \Sigma K(G,n)] \xleftarrow{\Sigma} [S^{i-1}, K(G,n)] \xrightarrow[\cong]{f \circ} [S^{i-1}, \Omega K(G,n+1)] \xrightarrow[\text{1:1}]{(\epsilon \circ) \circ \Sigma} [S^i, K(G,n+1)] \end{equation*} where $\Sigma$ denotes the suspension maps and $\epsilon$ is the counit of the adjunction.

By the Freudenthal suspension theorem, the first map is an isomorphism when $i < 2n$, and thus the composite is a bijection of sets at least when $i < 2n$.

The composite takes an element $\alpha \in \pi_i(\Sigma K(G,n))$ to the class of $\epsilon \circ \Sigma(f \circ \Sigma^{-1} \alpha) = (\epsilon \circ \Sigma f) \circ \alpha$. Since $(\epsilon \circ \Sigma f)$ is the adjoint map to $f$, this composite is precisely the induced homomorphism $\pi_i(\Sigma K(G,n)) \to \pi_i(K(G,n+1))$. Hence this homomorphism is an isomorphism when $i < 2n$.

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I don't understand why the answers above are so long.

For any CW complex $X$ based at $x$ you have the evaluation map $\Sigma \Omega X \to X$ defined by sending $\gamma \wedge t \to \gamma(t)$ where $\gamma$ is a loop in $X$ based at $x$.

Now let $X=K(G,n+1)$.

We then have the map $\Sigma K(G,n) = \Sigma \Omega K(G,n+1) \xrightarrow{\text{ evaluation}} K(G,n+1)$.

The only fact that I have used is that $\Omega K(G,n+1)= K(G,n)$. All the equal signs are homotopy equivalences.


edit: To solve a simpler but closely related problem, consider the thom spectra $MSO(k)=$ thom space of the tautological bundle over $BSO(k)=$an oriented grassmanian. The spectral 'evaluation' map $\Sigma MSO(n) =MSO(n) \wedge S^1 =MSO(n) \wedge M(SO(1)) \to M(SO(n) \oplus SO(1)) \to M(SO(n+1))$ induces an isomorphism in $\pi_j$ for $j<2k$. This can be seen by showing the isomorphism on $H^j$ and using the thom isomorphism theorem. Details are given in page 537 in Tom Dieck's Algebraic Topology Book.(I will write it down here later when I have time). Thus this quantifies the degree to which this map is a homotopy equivalence, for if all the homotopy groups induced by this map were the same, the evaluation map would be a homotopy equivalence, since the associated bundle is a CW complex up to homotopy.

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  • $\begingroup$ The question isn't just what the map is (everyone can see it's the adjunct of the map $K(G,n) \to \Omega K(G,n+1)$...), but what its behavior in homotopy is. OP was under the impression that $\Sigma K(G,n)$ was a $K(G,n+1)$, which isn't true. $\endgroup$ – Najib Idrissi Feb 29 '16 at 8:24

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