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First, let $\mathcal{G}$ be a groupoid. Then an automorphism $\gamma\colon X\rightarrow X$ in $\mathcal{G}$ considered as a loop in the nerve of $\mathcal{G}$ is homotopic to the point $X$ if and only if $\gamma=\mathrm{id}_X$.

This is no longer true in an arbitrary category $\mathcal{C}$. Consider for example the category consisting of two objects $X,Y$, an automorphism $\gamma\colon X\rightarrow X$ with $\gamma^{-1}=\gamma$ and another arrow $\alpha\colon X\rightarrow Y$ such that $\alpha\circ\gamma=\alpha$, i.e. $\alpha$ (co)absorbs $\gamma$. Surely, we then have the homotopy $$\xrightarrow{\gamma}\ \sim\ \xrightarrow{\gamma}\xrightarrow{\alpha}\xleftarrow{\alpha}\ \sim\ \xrightarrow{\alpha\circ\gamma}\xleftarrow{\alpha}\ \sim\ \xrightarrow{\alpha}\xleftarrow{\alpha}\ \sim\ X$$ Similarly, if we reverse the direction of the arrow $\alpha$, we have an arrow which absorbs $\gamma$.

Question: Can you provide an example of a category with a non-trivial automorphism which is neither absorbed nor co-absorbed by another arrow but is null-homotopic.

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2 Answers 2

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How about adding another object $W$ to your example with an automorphism $\delta:W\to W$ with $\delta^{-1}=\delta$ and arrows $\theta,\phi:X\to W$ with $\delta\circ\theta=\phi=\theta\circ\gamma$?

Then $\delta$ is null-homotopic but neither absorbed nor co-absorbed by any arrow:

Writing $\psi'$ for the path going backwards along an arrow $\psi$:

$$\delta\sim\delta\theta\theta'\sim\theta\gamma\theta'\sim\theta\operatorname{id}_X\theta'\sim\theta\theta'\sim\operatorname{id}_W$$

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  • $\begingroup$ Thanks for the answer. Could you tell me why $\delta$ is null-homotopic? Maybe I'm just blind. $\endgroup$
    – Werner Thumann
    Commented May 1, 2014 at 15:51
  • $\begingroup$ @WernerThumann: I've edited my answer with an explanation. $\endgroup$
    – Jeremy Rickard
    Commented May 1, 2014 at 16:01
  • $\begingroup$ @WernerThumann: Basically, you took a category with a non-trivial loop and coned it off by adjoining a terminal object. I've taken the cylinder and coned off one end. $\endgroup$
    – Jeremy Rickard
    Commented May 1, 2014 at 16:04
  • $\begingroup$ Ah, $\gamma$ is still absorbed by $\theta$ (and therefore by $\phi$). Thanks for the example. Btw, the order of the arrows was better before your edit. Path concatenation usually is written from left to right. I always prefer to write $ab$ instead of $b\circ a$ in private so that I don't get confused all day. ;) $\endgroup$
    – Werner Thumann
    Commented May 1, 2014 at 16:14
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One can in fact construct cancellative categories all of whose endomorphisms are automorphisms and for which some automorphisms are null homotopic in the nerve. For a cancellative category no arrow is absorbed by any other.

One can get an example by taking a Clifford 0-E-unitary inverse semigroup $S$ which is not strongly 0-E-unitary and take the subcategory of the idempotent splitting of $S$ with objects the nonzero idempotents and arrows the split epis.

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