Group Cohomology as Cohomology of the Grothendieck Topos BG

Group Cohomology as Cohomology of the Grothendieck Topos $BG$$BG$

Let $G$$G$ be a discrete group. Consider its classifying topos

$$BG:=G\text{-}\mathbf{Set}\simeq \mathrm{Fun}(\mathbf{B}G,\mathbf{Set}),$$

where $\mathbf{B}G$$\mathbf BG$ is the one-object groupoid with automorphism group $G$$G$.

The first observation is that linear algebra internal to $BG$$BG$ is precisely representation theory. If $k$$k$ is given the trivial $G$$G$-action, then

$$k\text{-}\mathrm{Mod}(BG)\simeq {\mathrm{Rep}}_k(G)\simeq k[G]\text{-}\mathrm{Mod}.$$

In other words, what externally looks like a $G$$G$-representation is simply an internal $k$$k$-linear space in the topos $BG$$BG$: the $G$$G$-symmetry has been absorbed into the ambient universe.

Now consider the global section functor

$$\mathrm{\Gamma }(BG,-)={\mathrm{Hom}}_{BG}(1,-).$$

For an internal $k$$k$-module $M$$M$, a global section $1\to M$$1\to M$ is precisely an element fixed by every $g\in G$$g\in G$. Hence

$$\mathrm{\Gamma }(BG,M)=M^G.$$

But the invariant functor $(-{)}^G$$(-)^G$ is left exact, and its right derived functors are exactly group cohomology:

$$H^n(G,M)=R^n(-{)}^G(M).$$

Therefore

$$H^n(G,M)=R^n\mathrm{\Gamma }(BG,M)=H^n(BG,M).$$

So group cohomology is literally the sheaf cohomology of the topos $BG$$BG$.