3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ What is $\hom(p_2,p_1)$? $\endgroup$ Apr 11, 2013 at 21:48
  • $\begingroup$ $hom(p_2, p_1)$ is empty. Sorry, I forgot to say that. $\endgroup$ Apr 11, 2013 at 22:01

1 Answer 1

6
$\begingroup$

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 $id_\mathcal{C} \Rightarrow inc \circ F$.

$\endgroup$
2
  • $\begingroup$ Don't you need that the action of $G$ on $X$ is free to get the functor $F$? $\endgroup$ Apr 11, 2013 at 21:03
  • 1
    $\begingroup$ Ah, but up to homotopy I can replace X by G. $\endgroup$ Apr 11, 2013 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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