# If a functor commutes with all filtered colimits, then does it commute with all existing colimits?

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 forgetful functor from rings to sets commutes with filtered colimits, but not with all colimits. Commented Jul 26, 2023 at 16:55

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 $$\operatorname{colim} F(J) = \operatorname{colim} F(K)$$.
• Yeah...It does not hold. It holds only if $F$ commutes with $J$, then this is circular. Thank you very much^^ Commented Jul 26, 2023 at 17:14