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Let $\mathcal{C}$ be a small category. Giving a morphism $u: X \to Y$ in $\mathcal{C}$ is equivalent to giving a functor $f: \{0 < 1 \} \to \mathcal{C}$ (with $f(0) = X$, $f(1) = Y$, and $f(0 \to 1) = u$). Accordingly, such a morphism induces a path $Bf: I = B\{0 < 1 \} \to B\mathcal{C}$ from the 0-cell $X$ to the 0-cell $Y$. Is the converse true? That is,

If there exists a path between 0-cells in $B\mathcal{C}$, does there exist a corresponding morphism in $\mathcal{C}$?

(Here $B \mathcal{C}$ denotes the geometric (or topological) realization of $\mathcal{C}$.)

Edit: Sorry, my language was imprecise the first time round. I am not asking if all paths are induced $Bf$ for some $f$ as above. My question is

If two 0-cells are path-connected in the realisation, then does there exist a morphism in the category from one of the corresponding objects to the other?

So, if a path-component contains both 0-cells $X$ and $Y$, does there exist a morphism $X \to Y$ or $Y \to X$?

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There are lots of contractible categories. – Martin Brandenburg Mar 17 '13 at 22:18
Question edited. – Joshua Seaton Mar 17 '13 at 22:26
A sequel question:… – Joshua Seaton Mar 18 '13 at 0:45
@vonbrand: that was a dumb edit, and it was pathetic you felt the need to do that. – Joshua Seaton Mar 18 '13 at 1:26
up vote 3 down vote accepted

No, of course not: once we pass to $\textbf{Top}$ we lose all sense of directionality. Let $\mathcal{C} = \{ 0 \to 1 \}$. Then there is a path $[0, 1] \to B \mathcal{C}$ corresponding to the morphism $0 \to 1$, but its inverse does not correspond to any morphism in $\mathcal{C}$.

The answer to the revised question is still no. Let $\mathcal{C} = \{ Y \rightarrow X \leftarrow Z \}$. In $B \mathcal{C}$, there is a path from $Y$ to $Z$, but there is no morphism $Y \to Z$ or $Z \to Y$ in $\mathcal{C}$.

What is true is the following: if $\mathcal{C}$ is a groupoid, then the homotopy type of $B \mathcal{C}$ is enough to determine $\mathcal{C}$ up to category-theoretic equivalence. This is the 1-dimensional case of the homotopy hypothesis.

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(The notion of classifying diagram, due to Rezk, is an attempt to rectify this inadequacy: instead of taking $B \mathcal{C}$ outright, we instead look at the simplicial space whose $n$-th level is $B (\operatorname{iso} \mathcal{C}^{[n]})$.) – Zhen Lin Mar 17 '13 at 22:19
Sorry, Zhen. I wasn't precise. I've fixed the wording of the question. Please address it when you get the chance. Thanks anyway for the response. – Joshua Seaton Mar 17 '13 at 22:29
Can one say anything then if two 0-cells lie in the same path-component? Maybe that there exist a chain of intermediate $Z$'s as above? – Joshua Seaton Mar 17 '13 at 23:39
It should be true that any two 0-cells in the same path component of $B \mathcal{C}$ should be connected by a zig-zag of arrows in $\mathcal{C}$, but the homeomorphism type of $B \mathcal{C}$ is a bit too complicated for me to check this directly. – Zhen Lin Mar 17 '13 at 23:49

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