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?

  • $\begingroup$ Welcome to Mathematics Se @f31 $\endgroup$ – magma Nov 11 '12 at 18:40

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 $\mathcal{C}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.