MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $


share|cite|improve this question
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. – Eric Wofsey Feb 22 '13 at 19:26
up vote 4 down vote accepted

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.

share|cite|improve this answer
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. – Omar Antolín-Camarena Feb 22 '13 at 21:28

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.

share|cite|improve this answer
I like the notation $\pi_{\le 1}(X)$ for the fundamental groupoid of $X$! Very descriptive. Is this standard notation? – Mark Grant Feb 23 '13 at 7:43
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? – Ronnie Brown Feb 23 '13 at 11:05
@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 $\pi_{\le 1}(X)=\pi_1(X,X)$. – Omar Antolín-Camarena Feb 23 '13 at 12:56
@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. – Ronnie Brown Feb 24 '13 at 10:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.