I am trying to understand the topologycal interpretation of group chomomology. I am familiar with the algebraic definition of cohomology for a group $G$, i.e. $H^i(G; A) := \text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$.
I have read on the internet that this should be tha same as the singular or cellular cohomology of a $CW-$complex $K(G,1)$, but this is not clear to me. Wherever I found this assertion, they would do the following construction:
Let $Y$ be a cellular $K(G,1)$ and let $\widetilde{Y}$ be its universal covering space. I can lift $Y$'s cell structure to $\widetilde{Y}$, then $G$, being the fundamental group of $Y$, will permute freely the cells of $\widetilde{Y}$ and the covering map will be exactly the projection on the orbit space. So the augmented cellular chain complex of $\widetilde{Y}$ is a free $G-$complex, which is exact because $\widetilde{Y}$ is contractible. Thus, this complex is a free resolution of $\mathbb{Z}$ over $\mathbb{Z}G$, which I can use to compute $\text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$.
My problem is that, from this construction, it seems to me that I would be getting the cohomology of $G$ not as the cellular cohomology of the $K(G,1)$ $Y$, but as the cellular cohomology of its universal cover $\widetilde{Y}$.
Where am I mistaken? Is the construction for getting a free $G$-resolution of not the one I encountered? Do you have any references where they prove explictly why the cohomology of $G$ is exactly the same as the cellular cohomology of a $CW-$complex $K(G,1)$, or could you explain to me how this works?

  • $\begingroup$ My suggestion is to read Brown's book, he gives both algebraic and topological viewpoints on group cohomology. $\endgroup$ May 7, 2021 at 22:17

1 Answer 1


You're skipping the step where you go from the free resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ to actually computing the Ext. The free resolution is the cellular chain complex of $\widetilde{Y}$. However, to compute $\text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$, you have to Hom this resolution into $A$ before taking cohomology. If $F$ is a free $\mathbb{Z}G$-module with basis $B$, then $\operatorname{Hom}_{\mathbb{Z}G}(F,A)$ is just $A^B$. Now, if $C_n(\widetilde{Y})$ denotes the $n$th cellular chain group of $\widetilde{Y}$, the $n$-cells of $\widetilde{Y}$ form a basis for $C_n(\widetilde{Y})$ over $\mathbb{Z}$. But over $\mathbb{Z}G$, a basis would instead be a set of representatives of the $G$-orbits of the $n$-cells of $\widetilde{Y}$. The $G$-orbits of the $n$-cells of $\widetilde{Y}$ just correspond to the $n$-cells of $Y$, so you can identify a basis for $C_n(\widetilde{Y})$ with the set of $n$-cells of $Y$. This means that when you Hom into $A$ over $\mathbb{Z}G$, you just get the group of functions from the set of $n$-cells of $Y$ to $A$, which is exactly the $n$th cellular cochain group of $Y$ with coefficients in $A$. So, the chain complex that computes $\text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$ has the same groups in it as the cellular cochain complex of $Y$ with coefficients in $A$. (Of course, you have to do some additional work to check that the differentials in these chain complexes are also the same.)

  • $\begingroup$ Thank you so much! Your answer was really useful, now understand $\endgroup$
    – abho
    May 7, 2021 at 15:13
  • $\begingroup$ I would like to ask one more question: if I take a $K(G,1)$ which is not a CW-complex, can I still say that it's cohomology equals that of $G$? I know that CW $K(G,n)$ are homotopy equivalent, but I have no clue about what is possible to say when the space isn't a CW-complex. $\endgroup$
    – abho
    May 15, 2021 at 15:42
  • 1
    $\begingroup$ Any two $K(G,n)$ spaces are still weakly homotopy equivalent and so their cohomologies are isomorphic. $\endgroup$ May 15, 2021 at 15:43
  • $\begingroup$ thanks for the answer! Could you maybe give any reference on a book where I might find that? I am studying mainly from Hatcher and I couldn't find anything there $\endgroup$
    – abho
    May 15, 2021 at 16:16
  • $\begingroup$ Every space is weak equivalent to a CW-complex. So if you have two $K(G,n)$s, they are each weak equivalent to CW $K(G,n)$s, but then those two CW $K(G,n)$s must be homotopy equivalent. $\endgroup$ May 15, 2021 at 16:25

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