This question is related to this question from Dinakar, which I found interesting but don't yet have the background to understand at that level.

Unless I'm mistaken, the rough statement is that $H^n(X;G)$ (the $n$-dimensional cohomology of $X$ with coefficients in $G$) should somehow correspond to (free?) homotopy classes of maps $X \to K(G,n)$. I want to understand this better, in relatively elementary terms. Here are some questions which (I hope) will point me in the right direction.

  1. What category are we working in? My guess is that $X$ should just be a topological space, the cohomology is singular cohomology, and our maps $X \to K(G,n)$ just need to be continuous.
  2. Does this carry over if we give $X$ a smooth structure, take de Rham cohomology, and require our maps $X \to K(G,n)$ to be smooth?
  3. How does addition in $H^n(X;G)$ carry over?
  4. How does the ring structure on $H^*(X;G)$ carry over? (This has probably been adequately answered to Dinakar already.)

3 Answers 3

  1. We are working in the homotopy category of topological spaces where morphisms are homotopy classes of continuous maps. More accurately, we tend to work in the based category where each object has a distinguished base point and everything is required to preserve that base point. The non-based category can be embedded in the based category by the simple addition of a disjoint base point, so we often pass back and forth between the two without worrying too much about it. The cohomology theory itself is slightly more interesting. For CW-complexes, it doesn't matter which one you choose as they are all the same. However, outside the subcategory of CW-complexes then the different theories can vary (as was mentioned in another question). So what we do is the following: using Big Theorems we construct a topological space, which we call $K(G,n)$, which represents the $n$th cohomology group with coefficients in $G$ for CW-complexes. So whenever $X$ is a CW-complex, we have a natural isomorphism of functors $\tilde{H}^n(X;G) \cong [X, K(G,n)]$, where the right-hand side is homotopy classes of based maps. For arbitrary topological spaces, we then define cohomology as $[X, K(G,n)]$. If this happens to agree with, say, singular cohomology then we're very pleased, but we don't require it.

  2. Depends what you mean by "smooth structure". Certainly in the broadest sense, you will get different answers if you insist on everything being smooth. But for smooth manifolds, continuous maps are homotopic to smooth maps (and continuous homotopies to smooth homotopies) so the homotopy category of smooth manifolds and smooth maps is equivalent to the homotopy category of smooth manifolds and continuous maps. However, you need to be careful with the $K(G,n)$s as they will, in general, not be finite dimensional smooth manifolds. However, lots of things aren't finite dimensional smooth manifolds but still behave nicely with regard to smooth structures so this shouldn't be seen as quite the drawback that the other answerants have indicated.

  3. Addition in $\tilde{H}^n(X;G)$ translates into the fact that $K(G,n)$ is an $H$-space. The suspension isomorphism, $\tilde{H}^n(X;G) \cong \tilde{H}^{n+1}(\Sigma X, G)$ implies the stronger condition that $K(G,n)$ is the loop space of $K(G,n+1)$ and so the $H$-space structure comes from the Pontrijagin product on a (based!) loop space. But the basic theorem on representability of cohomology merely provides $K(G,n)$ with the structure of an $H$-space.

  4. As for the ring structure, that translates into certain maps $K(G,n)\wedge K(G,m) \to K(G,n+m)$. I don't know of a good way to "see" these for ordinary cohomology, mainly because I don't know of any good geometric models for the spaces $K(G,n)$ except for low degrees. One simple case where it can be seen is in rational cohomology. Rational cohomology (made 2-periodic) is isomorphic to rational $K$-theory and there the product corresponds to the tensor product of vector bundles.

(It should be said, in light of the first point, that $K$-theory should only be thought of as being built out of vector bundles for compact CW-complexes. For all other spaces, $K$-theory is homotopy classes of maps to $\mathbb{Z} \times BU$.)

  • $\begingroup$ Thanks. So for question 3, we just identify the points of K(G,n) as loops in K(G,n+1), and then...what is the Pontrjagin product, exactly? To find the new image of a point of X, we take the product of its two old image points (i.e., compose the loops in K(G,n+1) that they represent)? $\endgroup$ Oct 28, 2009 at 6:06
  • $\begingroup$ The Pontrijagin product on a based loop space is concatenation of loops. This defines a morphism Omega Y x Omega Y -> Omega Y for any space Y. Then given two morphisms f,g X -> Y, we form the composition X -> X x X -> Omega Y x Omega Y -> Omega Y. $\endgroup$ Oct 28, 2009 at 7:43

Here's a precise statement: reduced singular cohomology $H^n(X;G)$ is naturally isomorphic to homotopy classes of pointed maps from $X$ to $K(G,n)$, for any pointed topological space $X$ having the homotopy type of a CW complex. Explicitly, the identity map $G = \pi_n(K(G,n)) = H_n(K(G,n); \mathbb{Z}) \to G$ gives an element $i_n$ of $H^n(K(G,n);G)$, and the isomorphism is given by taking a map $f : X \to K(G,n)$ to the class $f*(i_n)\in H^n(X;G)$.

By Yoneda, the additive and multiplicative structure on $H^*(X;G)$ come from certain (homotopy classes of) maps $K(G,n) \times K(G,n) \to K(G,n)$ and $K(G,n) \times K(G,m) \to K(G,m+n)$, respectively. The addition map is actually quite easy to see: $K(G,n)$ is the loopspace of $K(G,n+1)$, so it has a natural binary operation $K(G,n) \times K(G,n) \to K(G,n)$ given by concatenating loops. Since $K(G,n)$ is actually the double loopspace of $K(G,n+2)$, the Eckmann-Hilton argument (the same argument that shows higher homotopy groups are abelian) shows that this operation is commutative up to homotopy. I don't know of a good way to see the multiplication map.

As for your second question, the answer should be yes whenever it makes sense. For any good notion of a smooth structure, it should be true that smooth maps up to smooth homotopy are the same as continuous maps up to continuous homotopy (at least, it is true for smooth manifolds). However, as far as I know there is rarely a natural smooth structure to put on $K(G,n)$, so this doesn't make sense (though I may be wrong!). In particular, to do de Rham cohomology you presumably want $G$ to be $\mathbb{R}$ or $\mathbb{C}$, and then $K(G,n)$ is really monstrous geometrically. You may want to take a look at this question.


Indeed, the statement is that homotopy classes of continuous maps of pointed spaces $[X, K(G, n)]$ are in 1-1 correspondence with the elements of singular homology $H^n(X, G)$ for a CW-complex $X$.

The simplest example would be $G = \mathbb{Z}$, $n = 1$. Then you have $K(\mathbb{Z}, 1) = S^1$ and you can use a first cohomology class $c \in H^1(X, \mathbb{Z})$ to map the 1-skeleton of $X$ to $S^1$ (edge $e$ will make $c(e)$ loops around $S^1$). It's not hard to check that it gives the equivalence.

You also see that the choice of basepoint is irrelevant since you can shift it without affecting homotopy.

The example also helps to see that the additivity doesn't become obvious just from the things you wrote. To add properly, you need some kind of addition map on your target, that is, $K(G, n) \times K(G, n) \to K(G, n)$. How you prove this depends on your definition of Eilenberg-Mac Lane space, e.g. by universality. The ring structure comes from the wedge product (from an answer to this question).

If $X$ is smooth then the de Rham cohomology is the same as singular cohomology, but the space $K(G, n)$ has so little chance of being smooth (there was a question on MathOverflow explaining this) that the smooth maps are not really relevant, as expected for homotopy theory.

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    $\begingroup$ I don't like your summary of Rezk's answer to that question. It's an oversimplification to say that it "comes from the wedge product". It comes from the wedge product together with the properties of that with regard to connectivity together with the fact that the first homology and homotopy groups agree (modulo a little abelianisation). So I think best just to redirect to that question. $\endgroup$ Oct 28, 2009 at 7:45
  • $\begingroup$ Now, yes, sure, the fact follows not just from existence of some operation, but also from its properties. Could you say though how exactly the answer should be improved given that it already has the link? It's fine if you just edit it. $\endgroup$ Oct 28, 2009 at 20:10
  • $\begingroup$ E.g. the accepted answer says the same "As for the ring structure, that translates into certain maps K(G,n) wedge K(G,m) → K(G,n+m). " $\endgroup$ Oct 28, 2009 at 20:11

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