# Loopspace of Eilenberg Mac Lane space

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 $K(\mathbb{Z}_2 \times \mathbb{Z}_2, 1)$?

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If they were homotopy equivalent, they would have isomorphic fundamental groups, but by construction, their fundamental groups ($\mathbb Z/4$ and $\mathbb Z/2\times \mathbb Z/2$) aren't isomorphic... – Olivier Bégassat Jul 20 '12 at 22:50
Note I am asking about the respective loopspaces. I believe both of their loopspaces should have trivial fundamental groups. – Joe Jul 20 '12 at 22:51
Oh excuse me, I misread your question! – Olivier Bégassat Jul 20 '12 at 22:53
The following argument breaks down, because any old homotopy equivalence might not be of the form "$\Omega$ of a particular map between the $K(\dots,1)$". I still reproduce it here. $\pi_0(\Omega K(G,1))=[\mathbb S^0,\Omega K(G,1)]\simeq [\Sigma \mathbb S^0, K(G,1)]\simeq[\mathbb S^1,K(G,1)]=G$ is naturally a group, so the two spaces of path components (which on their own would be mere pointed sets) are naturally groups, and non isomorphic ones at that. A homotopy equivalence would yield an isomorphism between (the underlying pointed sets, but actually between) the group structures. – Olivier Bégassat Jul 20 '12 at 23:04
They are homotopy equivalent as abstract spaces (they are both discrete spaces with 4 points), but not homotopy equivalent as monoids or $A_\infty$ spaces. – Justin Young Jul 21 '12 at 19:21

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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Could you expand on the conclusion? Do you use some kind of CW-complex argument to reach the conclusion about homotopy equivalence at the end? (Edit) Ok thanks to Wikipedia :) – Olivier Bégassat Jul 20 '12 at 23:08

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 $\mathbb Z /4 \cong \mathbb Z/2 \oplus \mathbb Z/2$.

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