3
$\begingroup$

Suppose $C$ is a cocomplete category with initial object $I$ and terminal object $T\neq I$.

The forgetful functor $U:T/C\to C$ does not preserve colimits, because the empty colimit in the under category $T/C$ is its initial object $T$ and the empty colimit in $C$ is $I$.

My impression is that these are ''the only'' non-preserved colimits. Is this true? If not, does $U$ at least preserve pushouts?

  • 0
    Welcome to Mathematics Se @f312012-11-11

1 Answers 1

6

The dual claim is well-known: if $\mathcal{C}$ is a category and $X$ is any object in $\mathcal{C}$, then the forgetful functor from the slice category $(\mathcal{C} \downarrow X)$ to $\mathcal{C}$ preserves limits of all connected diagrams. Thus, the forgetful functor from the coslice category $(X \downarrow \mathcal{C})$ to $\mathcal{C}$ preserves colimits of all connected diagrams. The easiest way to see this is to check that wide pushouts and coequalisers are preserved.

In general the forgetful functor $(X \downarrow \mathcal{C}) \to \mathcal{C}$ will not preserve colimits of non-connected diagrams. For example, binary coproducts in $(X \downarrow \mathcal{C})$ will correspond to pushouts in $\mathcal{C}$.