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For some reason I convinced myself that a simplicial set (or maybe I mean directly Kan complex) is homotopy equivalent to the nerve of a groupoid if and only if it has no higher homotopy groups.

Is this true? (it seems like a proof could go through the fact mentioned in the title of my question)

And a related question: can a space with no higher homotopy groups be described (up to homotopy) as a CW complex with only 1-cells? (I expect not, is there a simple counterexample?)

The reason I ask is that it seems to me that CW complexes with only 1-cells seem special, in the sense that they do not have higher homotopy groups (right?) while something like $S^2$ has only 2-cells but has non-vanshing $\pi_3$.

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3 Answers 3

up vote 5 down vote accepted

Your first statement is true: there is an equivalence between homotopy 1-types and groupoids, given by taking the fundamental groupoid. (In general, one expects $n$-groupoids to be modeled by homotopy $n$-types).

An easy way to see this is as follows. There is an adjunction between simplicial sets and groupoids, where the left adjoint sends a simplicial set $X_\bullet $ to $\pi_1 X_\bullet$ (the groupoid, not the group) and where the right adjoint sends a groupoid to its nerve. For any simplicial set $X_\bullet$, there is an adjunction map

$$ X_\bullet \to N(\pi_1 X_\bullet)$$ which, roughly speaking, sends each point (vertex) in $X_0$ to the corresponding object in the groupoid, and sends each edge to the morphism in the fundamental groupoid that it determines. This is always an equivalence on $\pi_0$ and $\pi_1$. If there are no higher homotopy groups, it is an equivalence by Whitehead's theorem.

(In response to the title of your question: (nerves of) groupoids can be characterized as those simplicial sets with the unique horn filling property for all horns, just as categories can be characterized as those simplicial sets with the unique horn filling property for inner horns. If you believe the latter statement, it's not too hard to verify the former.)

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I just read your answer, is it obvious that the unit is an iso on $\pi_0$ and $\pi_1$? –  Jacob Bell Jun 7 '12 at 20:16
    
@donkeykong: It follows if you unwind the definitions. The adjunction map $X_\bullet \to N(\pi_1 X_\bullet)$ is an isomorphism on $\pi_0$ (it's an isomorphism on vertices) and is identified with the identity on $\pi_1$ by construction. –  Akhil Mathew Jun 7 '12 at 22:19
    
OK, I'll think about it, thanks. –  Jacob Bell Jun 7 '12 at 23:16

And a related question: can a space with no higher homotopy groups be described (up to homotopy) as a CW complex with only 1-cells? (I expect not, is there a simple counterexample?)

No. A (connected) CW-complex with only 1-cells (and 0-cells) is a (connected) graph, so homotopy equivalent to a wedge of circles (contract it along a spanning tree), so has fundamental group a free group by Seifert-van Kampen. So a $K(\pi, 1)$ where $\pi$ is not free is a counterexample (e.g. a compact orientable surface of genus at least $1$). Alternately, a CW-complex with only 1-cells (and 0-cells) has trivial $H_2$, which a $K(\pi, 1)$ need not have (the same counterexample suffices).

(Incidentally the second argument above proves that the fundamental group of a compact orientable surface of genus at least $1$ is not free. See this MO question for many other proofs of this.)

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thanks for that, that's a very good point. –  Jacob Bell Jun 7 '12 at 15:19

This is related to the notion of simplicial T-complex, for which there is a page http://ncatlab.org/nlab/show/T-complex

This is a simplicial set $K$ with in each dimension $>0$ a set $T_n \subseteq K_n$ of elements called thin and satisfying Keith Dakin's axioms:

  1. Degenerate elements are thin.

  2. Every horn has a unique thin filler.

  3. If all faces but one of a thin element are thin, then so also is the remaining face.

    A T-complex is of rank $n$ if $K_r=T_r$ for $r>n$. Then T-complexes of rank 1 are equivalent to groupoids; of rank 2 are equivalent to crossed modules over groupoids (Dakin); and Ashley's theorem is that T-complexes are equivalent to crossed complexes. For more see the link cited above.

In effect, T-complexes give a "linear" model of homotopy types, i.e. allowing operations of the fundamental groupoid, but no quadratic information such as Whitehead products. But as in mathematics generally, linear models can be useful!

The corresponding cubical T-complexes are also very useful.

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