I have a really tame category $C$: there are only finitely many objects $C_0$, each hom-set $C(x,y)$ for $x,y \in C_0$ has at most one element, and (aside from identity morphisms) if $C(x,y)$ is non-empty, then $C(y,x)$ is empty. I want to fatten this category a little without changing the homotopy type of the classifying space.

Let $\mu$ be a subset of the morphisms in $\bigsqcup_{x,y \in C_0}C(x,y)$. For each $f:x \to y$ in $\mu$, I want to add $g:y \to x$ subject to the relations $fgf = f$ and $gfg = g$. Call this new $g$-enhanced category $C[\mu]$. First question,

Is $C \to C[\mu]$ a standard operation and does it have a name? Clearly, this is similar to $-$ but not quite the same as $-$ localization around $\mu$.

Assume now that our selection of morphisms $\mu$ has been chosen so that there are no zig-zag cycles in $C$ of the form $$x_0 \to y_0 \leftarrow x_1 \to y_1 \to \ldots \leftarrow x_k \to y_k \leftarrow x_0,$$ where all the $x_j$ and $y_j$ are distinct, and only the forward maps lie in $\mu$. Thus, the addition of the corresponding $g$-morphisms does not destroy any existing loops in the classifying space $BC$. Second, and more important question,

Does the zig-zag acyclicity of $\mu$ suffice to guarantee homotopy equivalence $BC[\mu] \simeq BC$ of classifying spaces? If not, what more do we need from $\mu$ to achieve this homotopy equivalence?

  • $\begingroup$ I don't quite get this. If $f:x\to y$ is a morphism, then you have a zig-zag $x\rightarrow y\leftarrow x$ $\endgroup$ – Benjamin Steinberg Jul 19 '13 at 19:11
  • $\begingroup$ Okay, yes, I forgot to add that all x_j and y_j are distinct $\endgroup$ – Vidit Nanda Jul 19 '13 at 19:12
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    $\begingroup$ But if you start with the two-element poset $0<1$, which has a contractible classifying space, and do your construction with respect to the unique non-identity arrow, then don't you get something which is no longer contractible? $\endgroup$ – Benjamin Steinberg Jul 19 '13 at 19:14
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    $\begingroup$ @BenjaminSteinberg why isn't it contractible? $\endgroup$ – Vidit Nanda Jul 19 '13 at 19:59
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    $\begingroup$ I think the Quillen fiber over 0, in Benjamin Steinberg's example, has two objects and one morphism going between them, so it's contractible. The same is true for the Quillen fiber over 1. So it seems like Theorem A tells us that in the simplest example, the procedure in question doesn't change the homotopy type (hope I got this right...). $\endgroup$ – Dan Ramras Jul 19 '13 at 23:51

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