What are classifying spaces of algebraic categories like?

Question

Let $\mathcal{C}$ be a category. Recall that the nerve $N(\mathcal{C})$ of $\mathcal{C}$ is the simplicial set obtained by defining $N(\mathcal{C})_k$ to be the set of $k$-tuples of composable morphisms in $\mathcal{C}$, together with the obvious face and degeneracy maps.

To any simplicial set we can associate a CW complex by taking its geometric realization. Intuitively, this means we replace every element of $N(\mathcal{C})_k$ by a $k$-dimensional triangle (the case $k = 0$ being a point), and gluing them according to the face maps.

In particular, we may apply this procedure to the nerve of a category. I believe the resulting space is sometimes called the classifying space of your category. In some sense, one can think of the space as a tangible and geometric way to represent your category, especially if the category we care about is small. This leads me to the following question.

Question. What does the classifying space of a category $\mathcal{C}$ look like when $\mathcal{C}$ is an (essentially small) algebraic category? Is it interesting to look at? Can we describe its algebraic invariants, such as the fundamental group or homology groups?

For the sake of concreteness, let us specifically focus on the category of all finitely generated groups. But feel free to think about other cases, like the category of finite commutative rings. More geometrically minded people might want to think about the category of finite-type $S$-schemes or the category of smooth manifolds.

Note. The aforementioned categories are not small, though they are essentially small. Of course, one can simply take the skeleton of the essentially small category to obtain a small one, so that the resulting classifying spaces are actual sets.

What I tried. A simpler example would be to take the category of all finitely generated abelian groups, which has the advantage that they admit a nice classification. Writing down the definition of this 'space' is not too complicated, but I did not manage to find a way to simplify the construction, let alone think about more complicated categories like the ones described above.

Answer 1

One famous way to make this interesting is Quillen's $Q$-construction, which involves taking the classifying space of $Q\mathcal C$ instead of $\mathcal C$, where $\mathcal C$ is a sufficiently nice subcategory of an abelian category. $Q\mathcal C$ replaces the usual morphisms of $\mathcal C$ with "zigzags" or "spans" $A\leftarrow B\to C$, in which the leftward arrow is an epimorphism and the rightward arrow is a mono. Such a span identifies $A$ with a subquotient of $C$. In $Q\mathcal C$, an initial object $x$ must be uniquely a subquotient of every object. Since the only subquotient of $0$ is $0$, $x$ would have to be $0$, but $0$ admits at least two distinct morphisms to any nonzero object in $Q\mathcal C$: $0\leftarrow 0\to C$ and $0\leftarrow B\to B$. Similarly, $Q\mathcal C$ admits no terminal object. And indeed, in many interesting examples, $Q\mathcal C$ has a classifying space which is very far from contractible. Its homotopy groups are (identified with, after looping) the algebraic $K$-theory groups of $\mathcal C$, which are extremely important and difficult invariants. For instance, when $\mathcal C$ is finitely generated projectives over a ring $R$, $\Omega|Q\mathcal C|$ has homotopy groups beginning with the Grothendieck group $K_0(R)$ and the Bass-Schanuel group $K_1(R)=GL_\infty(R)/E(R)$, where $E(R)$ is the subgroup of elementary matrices in the infinite general linear group over $R$.

Answer 2

Eric Wofsey pointed out in the comments that the existence of an initial or final object ensures that the geometric realization will be contractible. I want to add an answer to show that (at least in hindsight) this is intuitively quite obvious.

Lemma. Let $N(\mathcal{C})$ be the nerve of a category $\mathcal{C}$. Suppose $\mathcal{C}$ has an initial object $x$. Then, for every map of simplicial sets $\partial \Delta^n \to N(\mathcal{C})$ such that the $0$-th vertex of $\partial \Delta^n$ gets sent to $x$, there exists an extension to a map $\Delta^n \to N(\mathcal{C})$.

Informally, this can be seen as follows. A map $\partial \Delta^n \to N(\mathcal{C})$ can be viewed as a choice of objects $x,y_1,\ldots,y_n$ in $\mathcal{C}$, together with maps $x \to y_i$ for all $i$, and maps $y_i \to y_j$ for all $i$ and $j$; the edges and higher-dimensional faces of $N(\mathcal{C})$ ensure certain commutativities between these maps. More precisely, it ensures that every chain $x \to y_{i_1} \to \cdots \to y_{i_k}$ of length less than $n$ will coincide with the map $x \to y_{i_k}$, and similarly that the chains $y_{i_1} \to \cdots y_{i_k}$ of length less than $n$ will coincide with the composition $y_{i_1} \to y_{i_k}$.

The existence of an extension to a map $\Delta^n \to N(\mathcal{C})$ essentially asks us whether this commutativity result still holds for chains of length $n$. Well, there's only one chain of this length, namely $x \to y_1 \to \cdots \to y_n$. Will the composition be equal to the given arrow $x \to y_n$? Well, yes, trivially so, because $x$ is initial! (And indeed, the fact that $x$ is initial is critical here! To convince yourself of this, try and find a counterexample for some low $n$, say $n = 2$ or $n = 3$, if $x$ is not initial.) Q.E.D.

Now take the geometric realization of the nerve $N(\mathcal{C})$. Regard the initial object as the base-point of your space. Geometrically speaking, what does a map $\partial \Delta^n \to N(\mathcal{C}))$ correspond to? It would be a map of spaces $S^n \to |N(\mathcal{C}|$. What, then, does the extension to a map $\Delta^n \to N(\mathcal{C})$ mean? It means that your map of spaces $S^n \to |N(\mathcal{C})|$ can be extended to a map $D^{n+1} \to |N(\mathcal{C})|$. Intuitively speaking, this means your spaces does not admit any holes, and since the spaces involved are all cell complexes, you can make precise that this is sufficient for your space to be contractible.