Given $A : \mathcal{C} \rightarrow \mathcal{E}$, when does $\operatorname{colim} A = A(1)$?

Question

Suppose $\mathcal{C}$ a small category and $A$ : $\mathcal{C} \rightarrow \mathcal{E} $ where $\mathcal{E}$ is a cocomplete category and $\mathcal{C}$ has a terminal object $1$

When is the colimit of $A$ equal to its image on the terminal object?

I thought something like:

If $A$ is a full functor then $\operatorname{im}(A)$ must be a full subcategory of $\mathcal{E}$ then the inclusion functor $\mathcal{I}$ must be final and $\operatorname{colim} (A \circ \mathcal{I})= \operatorname{colim} A$ and clearly $A(1)$ is the colimit of $A$ in $\operatorname{Im}(A)$

Is this right? Can we find weaker conditions ?

Answer 1

Suppose $\mathcal{C}$ is any category with a terminal object $1$ (not necessarily small) and $\mathcal{E}$ is any category (not necessarily cocomplete). Then for every functor $A : \mathcal{C} \to \mathcal{E}$, the colimit of $A$ exists and is canonically isomorphic to $A(1)$.

To see this, note first that there is a cocone of $A$ with object $A(1)$: indeed, for every $X \in \mathcal{C}$, let $t_X : X \to 1$ be the unique morphism, then $\varphi_X := A(t_X) : A(X) \to A(1)$ forms a cocone $\varphi$.

Now suppose $\psi = \{ \psi_X : A(X) \to U \}_{X \in \mathcal{C}}$ is another cocone of $A$, where $U \in \mathcal{E}$. Then $\psi_1 : A(1) \to U$ is a morphism such that $\forall X \in \mathcal{C}$, $\psi_1 \circ \varphi_X = \psi_X$ (because $\psi$ is a cocone and $\phi_X = A(t_X)$). Conversely, if $f : A(1) \to U$ is another morphism such that $f \circ \varphi_X = \psi_X$, then in particular $f \circ \varphi_1 = \psi_1$; but by definition $\varphi_1 = A(t_1) = A(\operatorname{id}_1) = \operatorname{id}_{A(1)}$, hence $f = \psi_1$. This proves the universal property: $$A(1) \cong \operatorname{colim} A.$$