# Group cohomology as cohomology of a K(G,1) CW-complex

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?

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

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.)

• Thank you so much! Your answer was really useful, now understand
– abho
May 7, 2021 at 15:13
• 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.
– abho
May 15, 2021 at 15:42
• Any two $K(G,n)$ spaces are still weakly homotopy equivalent and so their cohomologies are isomorphic. May 15, 2021 at 15:43
• 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
– abho
May 15, 2021 at 16:16
• 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. May 15, 2021 at 16:25