criteria for equivalence to the trivial category $\ast$ I'm trying to determine when a category is equivalent to the category $\ast$ - the category with one object and one morphism. Such category clearly cannot have a morphism which is not an isomorphism, and I think that the condition $|Hom_C (X,Y)|=1$ for every two objects $X,Y$,  is also needed. Am I right? how can I prove a necessary and sufficient condition?
 A: A category $C$ is equivalent to the trivial category (which I'll denote by $1$, with unique object $*$ and unique arrow $\text{id}_*$) if and only if it is nonempty and $|\text{Hom}(X,Y)| = 1$ for all objects $X$ and $Y$. I'll give you a concrete proof straight from the definitions. 
Proof: Note that $1$ is the terminal category: for any category $C$, there is a unique functor $!_C\colon C\to 1$ sending every object to $*$ and every arrow to $\text{id}_*$. In particular, $!_1 = \text{Id}_1$. 
Now suppose $C$ is nonempty and $|\text{Hom}(X,Y)| = 1$ for all objects $X$ and $Y$ in $C$. Let's note that every arrow in $C$ is an isomorphism. Indeed, if $f\colon X\to Y$ is an arrow, then letting $g\colon Y\to X$ be the unique arrow in $\text{Hom}(Y,X)$, we have $g\circ f = \text{id}_X$ and $f\circ g = \text{id}_Y$, since these are the unique arrows in $\text{Hom}(X,X)$ and $\text{Hom}(Y,Y)$, respectively. 
Since $C$ is nonempty, we can pick an object $X$ in $C$. Define a functor $F\colon 1\to C$ by $F(*) = X$ and $F(\text{id}_*) = \text{id}_X$. We need to show that $F$ and $!_C$ form an equivalence of categories. We have $!_C\circ F  = \text{Id}_1$, since this is the unique functor $1\to 1$. It remains to define a natural isomorphism $\eta\colon \text{Id}_C \cong (F \circ !_C)$. For each $Y\in C$, $F(!_C(Y)) = F(*) = X$. Let $\eta_Y\colon Y\to X$ be the unique arrow in $\text{Hom}(Y,X)$, which is an isomorphism. All naturality squares automatically commute, by our assumption on uniqueness of arrows. So we're done.
Conversely, suppose $C$ is equivalent to $1$. Then there is a functor $F\colon 1\to C$ which forms an equivalence of categories with the unique functor $!_C\colon C\to 1$. Let $X = F(*)$, and note that the existence of $X$ shows that $C$ is nonempty.
Now we have a natural isomorphism $\eta\colon \text{Id}_C\cong (F\circ !_C)$, with component isomorphisms $\eta_Y\colon Y\to X$ for all objects $Y$ in $C$. In particular, $\text{Hom}(Y,Z)$ is nonempty for all $Y$ and $Z$, since it contains $\eta_Z^{-1}\circ \eta_Y$. It remains to show that for any $f \in \text{Hom}(Y,Z)$, $f = \eta_Z^{-1}\circ \eta_Y$. By naturality, $$\eta_Z\circ f = F(!_C(f))\circ \eta_Y = F(\text{id}_*)\circ \eta_Y = \text{id}_X\circ \eta_Y = \eta_Y.$$ So $f = \eta_Z^{-1}\circ \eta_Y$, as desired. 
A: The functor $\mathbf{Cat} \to \mathbf{Set}$ that sends a small category to its set of objects has a right adjoint $R$ that sends a set $X$ to the category whose set of objects is $X$, set of morphisms is $X \times X$, domain map is the projection to the first coordinate, codomain map is the projection to the second coordinate, identity-assigning map is the diagonal map, and composition law is $(y,z) \circ (x,y)=(x,z) \, \forall x,y,z \in X$. A small category $C$ is equivalent to the trivial category with one object and one morphism if and only if $C$ is isomorphic (strictly) to the category $R(X)$ for some nonempty set $X$.
A: Remark first that a category $\mathbf C$ as a terminal object if and only if the functor $!:\mathbf C \to \mathbf 1$ has a right adjoint.
Then, $!:\mathbf C \to 1$ is an equivalence if and only if it has a right adjoint $R$ such that $R \circ {!} \simeq \mathrm{id}_{\mathbf C}$. By the previous remark, the latter condition is equivalent to $\mathbf C$ having a terminal object which is isomorphic to every object in $\mathbf C$, which in turn is equivalent to $\mathbf C$ being non empty and having exactly one morphism between each pair of objects.
