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Let $\mathcal{C}$ be an ordinary 1-category and suppose that there exists some object $X \in \mathcal{C}$ such that the following conditions are satisfied,

(1) For every $C \in \mathcal{C}$ we have $\operatorname{Hom}(X,C)\neq \emptyset$ , $\operatorname{Hom}(C,X)\neq \emptyset$.

(2) $\operatorname{Hom}(X,X)=*$.

It is easy to see that its 1-groupoidification must be categorically equivalent to the point.

I would like to know if these 2 conditions already imply contractibility of the nerve of $\mathcal{C}$.

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    $\begingroup$ By groupoidification do you mean the "core" where you take the isomorphisms that already exist or the "free groupoid" which freely adds inverses to all morphisms (the right and left adjoints respectively to the inclusion of groupoids into categories)? $\endgroup$
    – Max New
    Commented Dec 10, 2019 at 15:55
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    $\begingroup$ To make @MaxNew's objection more precise: on the core you know that $\operatorname{Aut}(X)$ is still trivial by (2), but you don't know if the category is connected. On the free groupoid you know connectedness by (1), but $\operatorname{Aut}(X)$ could be bigger now. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 11, 2019 at 2:37
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    $\begingroup$ My apologies for the late reply. By 1-groupoidification I mean the free groupoid, i.e. the truncated version of the Kan fibrant replacement. Using the description of the Hom-sets in terms of zig-zags of morphisms one can simplify each zigzag with source and target X to obtain the identity if I am not mistaken. $\endgroup$
    – F.Abellan
    Commented Dec 11, 2019 at 11:25

2 Answers 2

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Let $\mathcal{C}$ be the category with two objects $X$ and $Y$, generated by morphisms $\alpha_1,\alpha_2:X\to Y$ and $\beta_1,\beta_2:Y\to X$ subject to relations $\beta_i\alpha_j=\text{id}_X$ for all $i,j$.

So the only non-identity morphisms other than $\alpha_1,\alpha_2,\beta_1,\beta_2$ are the compositions $\alpha_i\beta_j:Y\to Y$.

According to my calculations, the nerve is homotopy equivalent to a $2$-sphere, but here is a proof that at least it has the cohomology with coefficients in a field $k$ of a $2$-sphere, and is therefore not contractible.

I think it is a standard fact that the cohomology $H^n(B\mathcal{C},k)$ of the classifying space of $\mathcal{C}$ is equal to the extension group $\text{Ext}^n(\mathbf{k},\mathbf{k})$ in the category of functors from $\mathcal{C}$ to $k$-vector spaces, where $\mathbf{k}$ is the constant functor taking the value $k$.

For each object $V$ of $\mathcal{C}$, there is a projective functor $P_V$ whose value on an object $U$ is the vector space with basis $\mathcal{C}(V,U)$, and a morphism $\alpha:V\to V'$ induces a morphism of functors $\alpha^\ast:P_{V'}\to P_V$ by composition.

A straightforward calculation shows that $$0\longrightarrow P_X\oplus P_X\stackrel{\pmatrix{\beta^\ast_1\\\beta^\ast_2}}{\longrightarrow} P_Y\stackrel{\alpha^\ast_1-\alpha^\ast_2}{\longrightarrow} P_X\longrightarrow\mathbf{k}\longrightarrow0$$ is a projective resolution of the constant functor, and applying the functor $\text{Hom}(-\mathbf{k})$ to the projective terms to calculate $\text{Ext}^*(\mathbf{k},\mathbf{k})$ gives $$k\stackrel{0}{\longrightarrow}k\stackrel{\pmatrix{1&1}}{\longrightarrow}k^2\longrightarrow0,$$ so $\text{Ext}^*(\mathbf{k},\mathbf{k})$ is one-dimensional in degrees zero and two, and zero in all other degrees.

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    $\begingroup$ Can you help us see why this gives a counterexample? $\endgroup$
    – Matt Feller
    Commented Dec 12, 2019 at 22:37
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    $\begingroup$ @JeremyRickard I will give it a look! Have you tried computing its homology? $\endgroup$
    – F.Abellan
    Commented Dec 13, 2019 at 8:24
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    $\begingroup$ @F.Abellan Its cohomology is that of a $2$-sphere. When I get the time, I'll add the calculation, unless somebody comes up with a quicker argument. $\endgroup$
    – Jeremy Rickard
    Commented Dec 13, 2019 at 8:53
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    $\begingroup$ Your second proposed counterexample, with W, doesn’t seem to satisfy condition (1) unless I’m misunderstanding something — it has no maps from W to X. (But your original counterexample sounds good to me — though it would be helpful if you could sketch the cohomology argument in the answer.) $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Dec 13, 2019 at 12:22
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    $\begingroup$ I'd be interested in how you argue for the homotopy type, but here's an argument from knowing the cohomology over fields: $H_\ast(BC)$ is a finitely-generated abelian group for each $\ast$, and $H_1(BC) = 0$. So it's easy to argue from the universal coefficient theorem using $k = \mathbb Q, \mathbb F_p$ that $H_2(BC) = \mathbb Z$. Since this is free, the universal coefficient theorem implies inductively that $H_\ast(BC) = 0$ for $\ast \geq 3$. So $BC = M(\mathbb Z, 2)$ is a Moore space and must be $S^2$ (explicitly, use Hurewicz to map in from $S^2$ and conclude by homology Whitehead). $\endgroup$
    – Tim Campion
    Commented Dec 16, 2019 at 18:08
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A counterexample is Connes’ cyclic category $\Lambda$: objects are $\langle n\rangle$ for all $n\in\mathbb N$; arrows $\langle n\rangle \to \langle m\rangle$ are homotopy classes of monotone, degree-$1$ maps $(S^1,\mu_{n+1})\to (S^1, \mu_{m+1}$) between circles with marked roots of unity. Its classifying space is known to be $\mathrm{B} S^1$ but it has an object $\langle 0\rangle$ satisfying your conditions because $\Lambda(\langle n\rangle, \langle 0\rangle ) \cong \Lambda (\langle 0\rangle, \langle n\rangle ) \cong \{0,\dots , n\}$ for all $n\in\mathbb N$

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