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Suppose I have a topological category $\mathcal{C}$ of the following form: The object space consists of just two points $p_1, p_2$. The endomorphism space of $p_1$ contains just the identity. The endomorphism space $End(p_2)$ of $p_2$ is a group $G$ and we have a restriction map $End(p_2) \to hom(p_1, p_2)$ which is a homotopy equivalence.

Using just this information, can we say anything about $B\mathcal{C}$?

My guess would be that it is either homotopy equivalent to $BG$ or contractible or something weird like $BG \times G$. Note that even though $hom(p_1,p_2)$ is a $G$-space, $B\mathcal{C}$ does not seem to be the bar construction $B(G,G,\ast)$ or is it? For the latter, the space of objects should be the $G$-space in question and not just two points. Moreover, note that $p_1$ is not necessarily an initial object, since the space $hom(p_1,p_2)$ is generally non-trivial. There also is an obvious map $BG \to B\mathcal{C}$, but I am unable to show that it is a homotopy equivalence, so maybe it isn't.

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What is $\hom(p_2,p_1)$? –  Fernando Muro Apr 11 '13 at 21:48
    
$hom(p_2, p_1)$ is empty. Sorry, I forgot to say that. –  Ulrich Pennig Apr 11 '13 at 22:01

1 Answer 1

up vote 6 down vote accepted

I think it is equivalent to $BG$. Picking an $x_0 \in X := hom(p_1, p_2)$ we get a map

$$X \to G$$

sending x_0 * g to g. Using this on morphism sets we get a map $F: \mathcal{C} \to G$ which you easily check to be a functor. Furthermore, the element $x_0$ can be used to get a natural transformation $id_\mathcal{C} \Rightarrow inc \circ F$.

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Don't you need that the action of $G$ on $X$ is free to get the functor $F$? –  Ulrich Pennig Apr 11 '13 at 21:03
1  
Ah, but up to homotopy I can replace X by G. –  Ulrich Pennig Apr 11 '13 at 21:22

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