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A very specific example: for $G=\mathbb{Z}$ and $n=m=1$, it is the map $S^1 \times S^1 \to CP^\infty$ which maps $S^1 \times S^1$ to the 2-skeleton $S^2$ of $CP^\infty$ by a degree 1 map. Here's a gen …

This is a reply to Alon's comment, but it's too long to be a comment and is probably interesting enough to be an answer. Here's an example Thom gives of a homology class that is not realized by a subm …

However, it is not an equivalence over $\mathbb{Z}$ because the cohomology of $BU$ is just a polynomial algebra and has no Steenrod operations. … Steenrod operations can be understood as obstructions to the cup product on ordinary cohomology being strictly commutative. …

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 … 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. …

All cohomology in this answer will have $\mathbb{Z}/2$ coefficients, and $K_n$ will denote $B^n(\mathbb{Z}/2)$. … By the Yoneda lemma, the cohomology $H^m(K_n)$ can also be thought of as the natural operations that take a cohomology class in degree $n$ on a space and give a cohomology class in degree $m$. …

Let $H\mathbb{Z}$ denote the spectrum for cohomology with coefficients in $\mathbb{Z}$, so your spectrum is $H\mathbb{Z}\wedge M_k(G)$. … Thus up to a degree shift, the cohomology theory $H\mathbb{Z}\wedge M_k(G)$ satisfies the Eilenberg-Steenrod dimension axiom, so it must coincide with $H^{*+k}(X,G)$. …

The group cohomology $H^n(G;A)$ can be identified with the cohomology $H^n(BG;A)$ of the classifying space $BG$ of $G$ with coefficients in A (interpreting the action of $G$, which is the fundamental group … The cup product on group cohomology is then the same as the usual cup product of singular cohomology. …

Let $M$ be a punctured torus and $N$ be a twice-punctured plane. Then $M$ and $N$ are homotopy equivalent, but their one-point compactifications are not (the first being a torus and the second having …

cohomology. … What is of interest is when you remove the dimension axiom to get "extraordinary" cohomology theories, which Oscar talks about in his answer. …

I believe it should be exactly the same as ordinary singular cohomology. … It should define a cohomology theory for the exact same reason that usual singular cohomology does (the usual proof of excision by subdivision seems to work since the cosubdivision of a $\Sigma_n$-invariant …

These are related to singular cohomology by looking only at constant sheaves. … It follows, for example, that for any CW-complex or any manifold, singular cohomology agrees with both Cech and derived cohomology of constant sheaves. …

Cup products give cohomology a natural graded ring structure, and the fact that this structure is preserved by continuous maps makes it often much easier to compute cohomology than homology. … Another (not unrelated) reason that cohomology can be easier to work with is that cohomology is a representable functor: H^n(X;A) is homotopy classes of maps from X to the Eilenberg-MacLane space K(A,n …