Is the loop space of the Eilenberg-MacLane space $K(G,1)$ dependent only on the cardinality of $G$? For instance, is the loop space of $K(\mathbb{Z}_4, 1)$ homotopy equivalent to that of $K(\mathbb{Z}_2 \times \mathbb{Z}_2, 1)$?
Loopspace of Eilenberg Mac Lane space
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1They are homotopy equivalent as abstract spaces (they are both discrete spaces with 4 points), but not homotopy equivalent as monoids or $A_\infty$ spaces. – 2012-07-21
2 Answers
Yes. Note that for a based space $(X, x_0)$, we have a fibration $\Omega(X, x_0) \hookrightarrow P(X,x_0) \longrightarrow X,$ where $\Omega(X,x_0)$ denotes the loop space of $X$ and $P(X,x_0)$ is the space of paths in $X$ based at $x_0$. Since $P(X,x_0)$ is contractible (shrink paths to the basepoint $x_0$), from the long exact sequence of the fibration we find that $\pi_k(X, x_0) \cong \pi_{k-1}(\Omega(X,x_0))$ for each positive integer $k$. So in our case, we have that $\pi_k(K(G,1)) \cong \pi_{k-1}(\Omega K(G,1))$ for each positive integer $k$. In particular, $\pi_0(\Omega K(G,1)) \cong G$ and $\pi_k(\Omega K(G,1)) \cong 0, ~k \geq 1.$
From the above, we see that $\Omega K(G,1)$ is a "$K(G,0)$." Although $K(G,n)$ is usually only defined for $n \geq 1$, we still get sensible spaces (homotopy equivalent to $G$ with the discrete topology) with similar properties in the case $n = 0$. In particular, we have $[X,K(G,0)] \cong \tilde{H}^0(X; G).$ Since $\Omega K(G,1)$ is a $K(G,0)$, $[\Omega K(G,1), K(G,0)] \cong \tilde{H}^0(\Omega K(G,1); G)$ and $[\Omega K(G,1), \Omega K(G,1)] \cong \tilde{H}^0(\Omega K(G,1); G),$ and therefore we can pick a map $f: \Omega K(G,1) \longrightarrow K(G,0)$ corresponding to the identity map on $\Omega K(G,1)$ via the above isomorphisms, which will induce isomorphisms on homotopy groups.
There is a theorem of Milnor saying that if $X$ has the homotopy type of a CW complex, then $\Omega X$ has the homotopy type of a CW complex. Therefore we have a map $f: \Omega K(G,1) \longrightarrow K(G,0)$ inducing isomorphisms on homotopy groups between spaces with the homotopy type of CW complexes. Therefore by the Whitehead theorem $\Omega K(G,1)$ is homotopy equivalent to $K(G,0)$, which is a discrete space of cardinality $|G|$.
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0Could you e$x$pand on the conclusion? Do $y$ou use some kind of CW-complex argument to reach the conclusion about homotopy equivalence at the end? (Edit) Ok thanks to Wikipedia :) – 2012-07-20
I think it's important to note that loop spaces have more structure than just their topology. Concantenation makes them into H-spaces ($A_\infty$ spaces), and if you use the Moore loop space, we can even make them into strictly associative monoids. J.P. May's book "The Geometry of Iterated Loop Spaces" basically says that every associative monoid ($A_\infty$ space) is equivalent as an associative monoid ($A_\infty$ space) to a loop space. Thus, the homotopy theory of loop spaces is essentially the same as the homotopy theory of associative monoids ($A_\infty$ spaces). Thus, when considering loop spaces we ought to consider the monoid structure as well. The monoids $G$ and $\Omega B G$ are weakly equivalent (here $BG = K(G, 1)$). So, $\Omega B G$ is really $G$ as a monoid, and $\Omega B\mathbb Z /4$ and $\Omega B(\mathbb Z /2 \oplus \mathbb Z/2)$ are not equivalent as monoids.
Edit: The key thing to note about maps of monoids, is that they induce maps of monoids on $\pi_0$, which any old continuous map will not do. If $\Omega B\mathbb Z /4$ and $\Omega B(\mathbb Z /2 \oplus \mathbb Z/2)$ were equivalent as monoids, then on $\pi_0$ we would get an isomorphism of monoids $\mathbb Z /4 \cong \mathbb Z/2 \oplus \mathbb Z/2$.