Suppose $F: C \to D$ is a functor between $C$ and $D$. If for all filtered diagram $I$, both the colimits $\mathsf{colim}F(I)$, $F(\mathsf{colim} I)$ exist and
$$ \mathsf{colim}F(I) = F(\mathsf{colim} I) $$ Then for any other arbitrary diagram $J$, if both the colimits $\mathsf{colim}F(J)$, $F(\mathsf{colim} J)$ exist, one always have
$$ \mathsf{colim}F(J) = F(\mathsf{colim} J) $$ ?
I got the following proof:
Proof: Since if $\mathsf{colim}J$ exists, adding $\mathsf{colim}J$ to the diagram $J$, i.e., considering the colimit cone, it's a filtered diagram. Denote this diagram as $K$, then from the following properties:
- cocones are filtered diagrams
- the (co)limit of a (co)cone is its apex
- the image of a (co)cone is still a (co)cone
$K$ is filtered diagram, $\mathsf{colim}K = \mathsf{colim} J$ and
$$ \mathsf{colim}F(J) = \mathsf{colim}F(K) = F(\mathsf{colim} K) = F(\mathsf{colim} J) $$
Is this argument correct? I am hesitant. It seems OK but too good to be true.
The statement is false. For example, the forgetful functor $U : \mathbf{Group} \to \mathbf{Set}$ preserves filtered colimits, but not binary coproducts for instance.
I think the invalid step is in the assertion that $\operatorname{colim} F(J) = \operatorname{colim} F(K)$.