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 $\mathcal{C}$.



About contractibility of certain categories
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.


