5
$\begingroup$

In "Higher algebraic K-theory I" Quillen defines a morphism inverting functor to be a functor from a category C to the category Sets which maps "arrows" in C to isomorphisms in Sets.

Proposition 1: The category of covering spaces of BC is canonically isomorphic to the category of morphism-inverting functors $F: C\rightarrow Sets$.

[For $C$ a small category, its classifying space $BC$ is the geometric realization of its nerve, $NC$]

This proposition plays an essential role in Quillen's Theorem 1 showing that his Q-construction agrees with Grothendieck's construction for $K_0$.

Theorem 1: $\pi_1(B(QC))$ is canonically isomorphic to the Grothendieck group $K_0(M)$

Questions: Have morphism-inverting functors played an important role in other contexts? Is there a more modern incarnation of morphism-inverting functors related to the fundamental groupoid of an infinity-category?

Improve this question
$\endgroup$
2
  • $\begingroup$ Infinity-categories have an underlying infinity-groupoid given by throwing out all non-invertible morphisms, but I wouldn't call it a fundamental groupoid. Rather, Joyal calls the infinity-groupoid of a topological space a "fundamental category". Also, the fundamental notion here is not the morphisms inverting functors, but the adjunction that I mentioned in my answer. $\endgroup$
    – Harry Gindi
    Commented Jan 28, 2010 at 17:33
  • $\begingroup$ I would suggest you read something like Goerss-Jardine and/or Joyal's lecture notes on quasicategories (available somewhere at nLab) if you want to get into higher categories in general. A really important concept that will not be in something as old as quillen is this notion of a simplicial presheaf. These have been used with great success by Voevodsky and others. I won't say that reading Quillen is a waste of time, but there are certainly more modern treatments of the material on which your time would be better spent. $\endgroup$
    – Harry Gindi
    Commented Jan 28, 2010 at 19:37

1 Answer 1

Reset to default
2
$\begingroup$

Proposition 1 is extremely straightforward to prove (provided you have some facts like the quillen adjunction between SSet and CGWH). Sing(|S|) gives you a simplicial set where all of the edges "forget" their direction, and when you apply the inverse of the nerve functor, you get back a copy of C with all of its arrows as isomorphisms. Covering spaces are equivalent to (etale) bundles (of sets) on a topological space, which by a theorem in Mac Lane (Sheaves in Geometry and Logic) is equivalent to taking sheaves on the space, so by unraveling these equivalences, you get your result. The last equivalence is probably one you're familiar with as the espace \'etal\'e. (While in general, the nerve functor does not have an inverse, the nerve of a category has some nice properties that make the total singular complex (the $Sing$ functor) pull back intact, modulo directedness of edges. If you think about the actual graph of the nerve of an ordinary category, it's not hard to see why this is true. This is precisely because the geometric realization "forgets" some information.)

The construction you're describing is generalized by a functor in HTT called the unstraightening functor, which you can read about in HTT Ch 2.2. With a number of more sophisticated results, we can generalize the adjunction between $Sing$ and $| \cdot |$ to a Quillen equivalence between SSet-Cat and CGWH-Cat.

HTT is Higher topos theory by J. Lurie.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I'll just motivate the proof sketch really quickly. Start out with a category C. Take |N(C)| : T. Consider now $Sing(T)$. It follows by some properties of the adjunction that $|Sing(T)|\cong T$. Then by some more abstract properties of the nerve, I can pass $Sing(T)$ back to $C':=N^{-1}(Sing(T))$. Also, $N(C')\cong Sing(T)$. This is all by formal nonsense involving adjunctions. Now taking sheaves on C', by transport of structure, gives us sheaves on T, which are equivalent to covering spaces. Similiarly, taking covering spaces on T gives us sheaves on C'. $\endgroup$
    – Harry Gindi
    Commented Jan 28, 2010 at 16:37
  • $\begingroup$ The reason we can pass freely between all of those adjunctions is because we've "restricted them to appropriate subcategories where the adjunctions restrict to equivalences. $\endgroup$
    – Harry Gindi
    Commented Jan 28, 2010 at 16:40
  • $\begingroup$ I just remembered, by the way, the adjoint to the nerve functor is the homotopy-category functor for simplicial sets, as defined in Lurie ch. 1.2.1, if I remember correctly. $\endgroup$
    – Harry Gindi
    Commented Jan 29, 2010 at 23:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .