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The classifying space $B\mathcal{C}$ of a small category $\mathcal{C}$ is by definition the geometric realization of the nerve of $\mathcal{C}$. Now let $\mathcal{C}_1$ and $\mathcal{C}_2$ be two small categories. Any functor $F : \mathcal{C}_1 \rightarrow \mathcal{C}_2 $ induces a continuous map $BF: B\mathcal{C}_1 \rightarrow B\mathcal{C}_2$.

My question is the following: when is a continuous map $f: B\mathcal{C}_1 \rightarrow B\mathcal{C}_2$ homotopic to a map of the form $BF$, for some functor $F:\mathcal{C}_1 \rightarrow \mathcal{C}_2 $

Thanks.

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    $\begingroup$ I don't know of any sort of general criterion, but my intuition is that typically the answer is "almost never". For instance, any finite CW-complex is homotopy equivalent to the nerve of a finite category; there can only be finitely many functors between two finite categories but there are often infinitely many homotopy classes of maps between two finite CW-complexes. This question is similar (and related) to asking when a map between simplicial complexes can be made into a simplicial map with respect to a fixed triangulation. $\endgroup$
    – Eric Wofsey
    Commented Feb 22, 2013 at 19:26

2 Answers 2

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Here's a counterexample. Let $M$ be a discrete monoid which is not a group. Consider the map $$ M = \hom (\Bbb N, M) \to \text{maps}_{*}(B\Bbb N , BM) = \Omega BM $$ ($\Omega BM =$ the based loops of $BM$ which is the same thing as Segal's group completion of $M$). This map is not a $\pi_0$ surjection since $M$ isn't a group.

Now, this fits into the context of your question since $\Bbb N$ and $M$ may be considered as categories with one object and a functor is just a homomorphism (i.e., $\cal C_1 = \Bbb N$ and ${\cal C}_2 = M$).

Remark: In the case when $\cal C_1$ and $\cal C_2$ are both groups, it would appear to me that the question you are asking has an affirmative answer.

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    $\begingroup$ The remark is correct: the homotopy category of 1-types is equivalent to the homotopy category of groupoids, so the answer is affirmative if both $\mathcal{C}_1$ and $\mathcal{C}_2$ are groupoids. $\endgroup$
    – Omar Antolín-Camarena
    Commented Feb 22, 2013 at 21:28
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Here's one instance where the answer is affirmative: when the target category is a groupoid (a little more general than my remark on John Klein's answer).

Let $\mathcal{C}$ be any category and $\mathcal{G}$ be a groupoid. Then $B\mathcal{G}$ is a 1-type (i.e., all $\pi_n$ with $n\ge 2$ vanish, with any base points), and therefore, for any space $X$, $\mathrm{maps}(X, B\mathcal{G})$ is a 1-type too, and in fact is just weakly homotopy equivalent to $B \mathrm{Fun}(\pi_{\le 1}(X), \mathcal{G})$ where $\pi_{\le1}(X)$ is the fundamental groupoid of $X$ (notice that since $\mathcal{G}$ is a groupoid, so is any functor category $\mathrm{Fun}(\mathcal{A}, \mathcal{G})$). Applying this to $X=B\mathcal{C}$, we get a weak homotopy equivalence $\mathrm{maps}(B \mathcal{C}, B \mathcal{G}) \simeq B \mathrm{Fun}(\pi_{\le 1}(B\mathcal{C}), B\mathcal{G})$. Now, $\pi_{\le 1}B \mathcal{C}$ is just $\mathcal{C}[\mathcal{C}^{-1}]$, the localization of $\mathcal{C}$ obtained by adding formal inverses for all morphisms of $\mathcal{C}$; and since $\mathcal{G}$ is a groupoid, any functor $\mathcal{C} \to \mathcal{G}$ automatically factors through $\mathcal{C}[\mathcal{C}^{-1}]$, so composition with the canonical functor $\mathcal{C} \to \mathcal{C}[\mathcal{C}^{-1}]$ induces an equivalence of categories $\mathrm{Fun}(\mathcal{C}, \mathcal{G}) \cong \mathrm{Fun}(\mathcal{C}[\mathcal{C}^{-1}], \mathcal{G})$. Equivalences of categories induce homotopy equivalences of classifying spaces, so we get a weak equivalence $\mathrm{maps}(B \mathcal{C}, B \mathcal{G}) \simeq B\mathrm{Fun}(\mathcal{C}, \mathcal{G})$, which is a little better than just saying that every function between the classifying spaces is homotopic to one coming from a functor.

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  • $\begingroup$ I like the notation $\pi_{\le 1}(X)$ for the fundamental groupoid of $X$! Very descriptive. Is this standard notation? $\endgroup$
    – Mark Grant
    Commented Feb 23, 2013 at 7:43
  • $\begingroup$ I'm not so keen on the notation $\pi_{\leqslant 1} X$ since I find the more general $\pi_1(X,A)$, the fundamental groupoid on a set $A$ of base points, a useful concept, where the set $A$ is chosen conveniently for the geometry of the situation, e.g. in van Kampen type problems, as introduced in my 1967 paper. The concept is also useful in discussing orbit spaces, where $A$ might be an orbit for the action, or, say, the set of all fixed points. Why are so many texts limited to the case $A$ is a singleton or $A$ is the whole space? $\endgroup$
    – Ronnie Brown
    Commented Feb 23, 2013 at 11:05
  • $\begingroup$ @MarkGrant: I think it's fairly standard. @RonnieBrown: I agree with the advantages of $\pi_1(X,A)$, but for the case $X=A$, I find $\pi_1(X,X)$ too long, and, of course most people would assume $\pi-1(X)$ is a group, so I use $\pi_{\le 1}(X)=\pi_1(X,X)$. $\endgroup$
    – Omar Antolín-Camarena
    Commented Feb 23, 2013 at 12:56
  • $\begingroup$ @Omar: Fair enough. I tend to think in terms of the difference between functors $\pi_1$ on spaces and on pointed spaces, the latter giving THE (in this case a sensible term) fundamental group. $\endgroup$
    – Ronnie Brown
    Commented Feb 24, 2013 at 10:47

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