Let $\mathcal{C}, \mathcal{D}$ be categories internal to topological spaces (or compactly generated Hausdorff spaces, if you like) $F,G\colon\mathcal{C}\rightarrow\mathcal{D}$ be continuous functors and $\eta\colon F\implies G$ a natural transformation, i.e. a continuous functor $$\mathcal{C}\times\{0,1\}\rightarrow \mathcal{D}.$$

Using the ordinary geometric realization, this functors induces a homotopy between $BF$ and $BG$ since $B$ preserves products and $B(\{0,1\})\cong[0,1]$.

If one uses the fat realization $B'(\_)$, products are probably preserved up to homotopy, but $B'(\{0,1\})$ will be something infinite dimensional. Is it still true that $B'F$ and $B'G$ are homotopic?

If the nerves of $\mathcal{C}$ and $\mathcal{D}$ are "good" in the sense of Segal, the ordinary realization coincides up to homotopy with the usual one and the question can be answered positively. However, is it still true in the "bad" world?

Edit: The fact that products are preserved up to homotopy is Proposition A.(iii) of Segal's "Categories and Cohomology Theories".

  • $\begingroup$ B'(2) is bipointed and contractible, so there's a path between the points. This gives you a homotopy (I've written 2={0<1} for where you wrote {0,1}, since it's not just a two-element set) $\endgroup$ Jul 10, 2015 at 15:43
  • $\begingroup$ There's still the issue of the product, so it's "almost gives you a homotopy" for now. $\endgroup$ Jul 10, 2015 at 15:48
  • 1
    $\begingroup$ This was proven by Segal (see my edit) hence the question can be answered affirmative. $\endgroup$
    – Simp
    Jul 10, 2015 at 16:44
  • $\begingroup$ I don't know why there is a vote to close this (I mean, in MO 1.0 we could close as "no longer relevant", but that's been deprecated). It seems like a good question, research level, that was answered, albeit quickly. $\endgroup$ Jul 11, 2015 at 4:32
  • $\begingroup$ @TheoJohnson-Freyd I can add an answer if desired... $\endgroup$ Jul 11, 2015 at 14:55

1 Answer 1


As pointed out by the OP, fat realisation preserves products up to homotopy, so we get a map $$B'(C)\times B'(\{0\lt1\}) \simeq B'(C\times \{0\lt1\}) \to B'(D).$$ The space $ B'(\{0\lt1\})$ is bipointed and path-connected (in fact contractible) so we get a homotopy $B'(C)\times [0,1]\to D$. Then one needs to show the maps it is a homotopy between are themselves homotopic to the two induced maps $B'(C\times \{i\})\to B'(D)$, $i=0,1$.


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