## Loopspace of an Eilenberg Maclane space K(G,n)

I've seen the fact that the loopspace $\Omega K(G,n)$ is homotopy equivalent to $K(G,n-1)$ mentioned in some places, but I have no idea why. Can anyone offer a good explanation? Also, what happens when $n=1$? Does that just mean the loopspace is a discrete space with $|G|$ points?

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Take a look at the "compact-open topology" section in the Munkres point-set topology text (or almost any other). Once you understand the result $A^{B^{C}}=A^{B \times C}$ for "reasonable spaces" as Blass mentions, it's a fairly standard exercise to show $\pi_0 \Omega X = \pi_1 X$, similarly $\Omega^i \Omega^j X \simeq \Omega^{i+j} X$, so $\pi_i \Omega X = \pi_{i+1} X$. – Ryan Budney Sep 30 2010 at 0:17
Re your last sentence: in this case the loop space is homotopy equivalent to a discrete space with |G| points. – Kevin Walker Sep 30 2010 at 1:07

In general,the map $P(X,x_0)\to X$ from the space of based paths in $X$ is a fibration with fiber $\Omega(X,x_0)$. Since $P(X,x_0)$ is contractible, by shrinking paths back toward the base point $x_0$, the homotopy long exact sequence of the fibration shows that $\pi_k(X,x_0)\cong \pi_{k-1}(\Omega(X,x_0))$. The result follows. And, yes, the loop space on a $K(G,1)$ is discrete, in one-to-one correspondence with $G$.

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For any (reasonable) space, the homotopy groups of $\Omega X$ are those of $X$ shifted by 1; $\pi_k(\Omega X)=\pi_{k+1}(X)$. Since the homotopy groups of $K(G,n)$ are $G$ in dimension $n$ and 1 in all other dimensions, the homotopy groups of $\Omega K(G,n)$ are $G$ in dimension $n-1$ and 1 in all other dimensions. That makes $\Omega K(G,n)$ a $K(G,n-1)$.

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If $X$ is $(n-1)$-connected, and $Y$ is a space with $\pi_n(Y) \cong G$ and no other nontrivial homotopy groups, then the map $[X, Y] \to \mathrm{Hom}(\pi_n(X), G)$ is a bijection (by elementary obstruction theory). It follows that if $Z$ is a space with $\pi_n(Y) \cong H$ and no other nontrivial homotopy groups, then $[Y, Z] \cong \mathrm{Hom}(G, H)$, and consequently, that if $G\cong H$, $Y$ and $Z$ are weakly equivalent.

Assuming they are CW complexes, they are homotopy equivalent; we call the common homotopy type $K(G,n)$.

So, we know that the loop space operator $\Omega$ has the effect of shifting homotopy groups downward. Thus $\Omega Y$ is a space with $\pi_{n-1}(Y)\cong G$ and all other homotopy groups trivial. Thus $Y \simeq K(G, n-1)$.

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I must confess to being a little dissatisfied with the other answers (so far). They all have a flavour of, "This is really quite simple, here's why." (I should make it clear that I do not infer anything as to the intent of the authors concerned). In fact, there's a lot of machinery in the background that makes these answers work. Someone working in algebraic topology has (probably) so internalised this machinery that they forget that someone coming to it afresh hasn't done so. The question, to me, reads as that of someone who hasn't yet internalised the machinery, so I'm going to give a slightly different answer.

The key piece of machinery comes in the question that should have been asked first (except that this isn't really a good MO question ..): "What is $K(G,n)$?". The true short answer is that it is a homotopy type. That is, it isn't really one thing, but an equivalence class of things defined by a common property:

A space (with the homotopy type of a CW-complex) is a $K(G,n)$ if it represents the reduced homology functor $\tilde{H}^n(-,G)$ on finite CW-complexes.

In truth, to represent the homology functor we must also include in to the data a particular class in $\iota \in \tilde{H}^n(K(G,n),G)$ so that the representing isomorphism

$$[X,K(G,n)] \to \tilde{H}^n(X,G)$$

is given by $\alpha \mapsto \alpha^* \iota$.

The first part of the Big Machine is called Brown representability. It says, in short:

There is a space which is a $K(G,n)$.

Now we do a bit of preparatory work. We can compute $\tilde{H}^n(S^k,G)$ quite simply to find that it is zero (by which I mean the trivial group) if $k \ne n$ and $G$ if $k = n$. By the isomorphism in Brown representability and the definition of homotopy groups, we deduce that if $X$ is a $K(G,n)$ then $\pi_k(X) = 1$ for $k \ne n$ and $\pi_n(X) = G$.

The next bit of the Big Machine is the following:

If $X$ and $Y$ are simply connected CW-complexes, then a weak equivalence is a homotopy equivalence.

(I want to attribute this to Whitehead, but my reference doesn't make any attribution so I shan't in case I'm wrong).

One very important thing to note here is that to apply this, one has to have a map. So saying $\pi_k(X) \cong \pi_k(Y)$ for all $k$ is not enough. These isomorphisms have to be generated by a map. But in our case, we do have such a map. If $X$ and $Y$ are $K(G,n)$s then $[X,X] \cong H^n(X,G) \cong [X,Y]$ so there is a map $X \to Y$ corresponding to the identity on $X$. As it corresponds to the identity on $X$, it's easy to show that it induces an isomorphism on homotopy groups. Hence (as $X$ and $Y$ are assumed to have the homotopy type of CW-complexes), $X \simeq Y$.

To apply all of this to the case in question, we need to prove two things:

1. If $X$ is a $K(G,n)$ then $\Omega X$ represents $H^{n-1}(X,G)$ (on finite CW-complexes).
2. If $X$ has the homotopy type of a CW-complex, then $\Omega X$ has the homotopy type of a CW-complex.

The second is a result due to Milnor. The first follows from the suspension isomorphism:

$$H^n(\Sigma Z, G) \cong H^{n-1}(Z, G)$$

If $X$ is a $K(G,n)$, then we get a natural isomorphism:

$$H^{n-1}(Z,G) \cong H^n(\Sigma Z, G) \cong [\Sigma Z, X] \cong [Z, \Omega X]$$

and hence $\Omega Z$ represents $H^{n-1}(-,G)$ on finite CW-complexes. Thus, by all the above, it is a $K(G,n-1)$ space and so is homotopy equivalent to any other $K(G,n-1)$ space.

Finally, I'd like to underline Kevin Walker's comment. If $G$ is discrete, $\Omega K(G,1)$ is homotopy equivalent to $G$, but may not be it exactly. A simple example is the case $G = \mathbb{Z}$. The circle $S^1$ is a $K(\mathbb{Z},1)$. $\Omega S^1$ is the space of all continuous maps from the circle to itself which fix the basepoint. That is most certainly not a discrete space! But it has components corresponding to $\mathbb{Z}$ (given by the winding number) and each component is contractible. Hence $\Omega S^1 \simeq \mathbb{Z}$. In fact, by interpreting $\Omega$ a little differently, you can get a situation in which $\Omega S^1 = \mathbb{Z}$, but to do this you need to interpret $\Omega$ as being polynomial loops.

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 @Andrew: The stated result (due to whitehead) does not need simple connectivity. Indeed, it is true for any pair of fibrant-cofibrant objects in a model category. The fibrant-cofibrant objects of $\operatorname{CGWH}$ with the Quillen model structure (weak equivalences = weak homotopy equivalences) are (if I remember correctly) CW complexes with no connectivity assumption. – Harry Gindi Sep 30 2010 at 9:02 With my "Algebraic Topology" hat on then I want to be sure that I don't say anything wrong. The book that I was checking my results from doesn't quite have the result that I needed, but had two results that chained together did, and one of them needed simply connected, so I put it in to be safe. – Andrew Stacey Sep 30 2010 at 9:41 Rereading this answer, I see that I've missed out lots of "tilde"s on my homology groups. All homology groups are reduced. (I'll fix this when I fix the simply-connected condition; I'm waiting to see what else I got wrong!) – Andrew Stacey Sep 30 2010 at 10:05

When I first saw this question I expected to find a number of votes to close it since the question asks about something that is surely covered in any algebraic topology textbook that includes basic homotopy theory. However, the question was not closed, and to my further surprise there have been some rather elaborate answers to so elementary a question.

Whether a simple answer is possible depends on which definition of a $K(G,n)$ one uses. Surely the simplest definition, and the one I thought was standard, given for example in Spanier, is that a $K(G,n)$ is a space whose homotopy groups are all trivial except for $\pi_n$ which is isomorphic to $G$. (When $n=0$ the definition needs to be adjusted slightly, or this case can be excluded entirely, as in Spanier.) With this definition the answer given by Allan Edmonds is perfectly satisfactory, using the path fibration, and so is the answer by Andreas Blass which is even more elementary, using just the definitions of the objects involved (and basic properties of function spaces).

Now, in practice one wants $K(G,n)$ spaces to be unique up to homotopy equivalence, and this requires restricting to spaces having the homotopy type of CW complexes. Uniqueness can then be proved by elementary arguments not using cohomology or obstruction theory. If one makes the requirement of having the homotopy type of a CW complex part of the definition of a $K(G,n)$ then to answer the original question one would have to quote Milnor's theorem that the loopspace of a CW complex has the homotopy type of a CW complex. This is not a trivial theorem, and it involves digressions into point-set topology. Milnor's theorem is surely a reassuring fact to know, but it can often be avoided by using CW approximations. Overall, it seems best not to require $K(G,n)$'s to have the homotopy type of CW complexes, although this may be a matter of personal preference.

As Andrew Stacey's answer shows, quite different definitions of $K(G,n)$'s are possible. These may have their virtues in certain contexts, but I would argue that when it comes to definitions, simplicity and minimality of prerequisites should be the highest priorities. With the simple definition of $K(G,n)$'s in terms of homotopy groups, the fact that they represent cohomology groups $H^n(X;G)$ of CW complexes $X$ as $[X,K(G,n)]$ then comes as a miraculous surprise, rather than something that is true by definition.

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 In so far as the first paragraph refers to me, had I come across this without any answers then I probably would have voted to close. But given that there were already answers, closing would have left those there and (as I said) I didn't like them. Also, in my own subject area, I feel less able to judge what should or shouldn't be closed because (as I hinted), I'm so far in that I don't really know where the boundary is. On to the mathematics. The question isn't really clear enough (which is what I was trying to get at) in that it doesn't explain what the questioner thinks K(G,n) is (ctd) – Andrew Stacey Oct 1 2010 at 7:27 (ctd) Again, had I been first on the scene I'd've asked for that clarification. The lack of article ("a K(G,n)" or "the K(G,n)") means that there are no clues either. So I would say that both of our answers can be summarised as: "If you actually define K(G,n) (rather than just having a vague notion of what it is), and then the rest is (fairly) automatic.". And so to the definitions. I find myself preferring to define something by what it does rather than what it is. So I'd rather that it was the simplicity of the homotopy groups of K(G,n) that came as a miraculous surprise. (ctd) – Andrew Stacey Oct 1 2010 at 7:30 (ctd) (Gosh, this is getting a bit long!). Defining K(G,n) as a space with certain homotopy groups makes me want to ask "So what?"! Why should I be interested in K(G,n) except as an exercise with which to torture algebraic topology students? But that it represents cohomology, ah, now that's something worth knowing! But then I've never taught algebraic topology and my head is full of cohomology operations, so I'm naturally inclined to that view and if I ever actually had to explain it all (in more than 2000 characters), I'm sure I'd end up agreeing with you! – Andrew Stacey Oct 1 2010 at 7:33