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It should send functors to continuous functions and natural transformations to homotopies.

I assume it should also send the interval category $0\rightarrow 1$ to the interval $[0,1]$ somehow.


I assume the codomain is $hTop$.

I don't know much about bicategories, but I can handle a tiny bit.

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The geometric realization of a category does what you describe.

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  • $\begingroup$ Thanks. Those definitions are a bit too much for me right now. I'm not very familiar with weighted colimits and all that. I'm gonna get there eventually. May I ask, simply for the category with two non trivial arrows $0\rightarrow 1\rightarrow 2$ what does its geometric realization look like? And if you have some other example you'd like to give, it's welcome as well. Cheers! $\endgroup$
    – Julián
    Apr 20 at 16:23
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    $\begingroup$ @Julián Here’s a description of how to construct the space for your category, which I hope will give an idea of how it works more generally. Start with a vertex for each object, so in this case three vertices: $1$, $2$, $3$. Add an edge from vertex $i$ to vertex $j$ for each non-identity morphism, so in this case three edges between pairs of vertices $(1,2)$, $(2,3)$, $(1,3)$. Glue the edges of a $2$-simplex (triangle) along the edges $(i,j), (j,k), (i,k)$ for each pair of composable morphisms $i\to j\to k$ and their composition $i\to k$, so in this case one $2$-simplex. $\endgroup$ Apr 20 at 20:14
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    $\begingroup$ @Julián So for your category you get a solid triangle. $\endgroup$ Apr 20 at 20:14
  • $\begingroup$ $1$-categories are always dimension $2$ or below? And in general $n$-categories are of dimension $(1+n)$ or below? ; could it be that taking border of the geometric realization is the same as the geometric realization of a "1 level truncation" (removing n cells of an n-category) ? If that's the case, I have a good intuition of how it works! $\endgroup$
    – Julián
    Apr 20 at 20:56
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    $\begingroup$ @Julián No, in general you carry on and adjoin a $3$-simplex for every set of three composable morphisms $i\to j\to k\to l$, and so on. But your example didn’t have three composable morphisms. $\endgroup$ Apr 20 at 22:31

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