Reducing the simplices in the nerve of a category with an object with trivial endomorphism monoid Let $C$ be a category with an object $X$ such that there are no non-trivial endomorphisms $X\rightarrow X$. Consider a simplex $\sigma$ of the nerve $NC$ of $C$. It is just a string of composable arrows in $C$. Define a $X$-component of $\sigma$ to be a maximal substring of the form $X\rightarrow X\rightarrow...\rightarrow X$. Let $N^r(C)$ be the simplicial subset of $N(C)$ consisting of the simplices with at most one $X$-component. Roughly speaking, you only consider those simplices where you travel over $X$ only once. For example, $A\rightarrow X\rightarrow B$ is allowed, but not $X\rightarrow A\rightarrow X$.
Question: Is the inclusion $N^r(C)\rightarrow N(C)$ a weak equivalence?
 A: I found a proof for this statement. But the reason is another one that one might think. Let $C^-$ be the full subcategory generated by the objects of $C$ minus the $X$. I claim: If there is a simplex with more than one $X$-component at all, then both the inclusions $NC^-\rightarrow NC$ and $NC^-\rightarrow N^rC$ are homotopy equivalences. I give a very short explanation. First observe that the existence of a simplex with more than one $X$-component implies the existence of an object $A$ and arrows $A\rightarrow X$ and $X\rightarrow A$. From this one can show that the comma category $C^-\downarrow X$ is filtered and therefore contractible. Quillen's A implies that the inlusion $NC^-\rightarrow NC$ is a homotopy equivalence. On the other hand $N^rC$ is the pushout of
$$NC^-\leftarrow N((C^-\downarrow X) * (X\downarrow C^-))\rightarrow \operatorname{Cone}N((C^-\downarrow X) * (X\downarrow C^-))$$
(That's in fact the reason why I defined $N^rC$). So also $NC^-\rightarrow N^rC$ is a homotopy equivalence.
Probably the following is the upshot of this question: Whenever you see a category with an object $X$ with no non-trivial endomorphisms but an object $A$ with $A\rightarrow X$ and $X\rightarrow A$, then you can kick the $X$ and the homotopy type of the category doesn't change. This follows easily, as pointed out above, from Quillen's Theorem A which we all love so much.
