Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 119 2 i Solution Created 2026-10-03 Updated 2026-10-05
Let be the given cocone under a diagram. For , choose an object of the comma category and defineSuch a choice exists because is a final functor. If gives a morphism , so , the cocone relation givesThe connected category condition makes the result independent of the choice along any zigzag.
For , choose ; then is available for , giving . Thus the maps form a cocone under a diagram on . Choosing shows that restriction recovers . Any extension must satisfy , proving uniqueness. This is cocone extension along a final functor.
If has a colimit, extend its universal cocone in this way. Every cocone on restricts to one on , factors uniquely through that colimit, and then agrees with its extension by uniqueness. Hence the same vertex is a colimit of . In particular,and existence of all -shaped colimits in implies existence of all -shaped colimits.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 119 3 Solution 2026-09-28
A cocone under a diagram with vertex is a natural family , so for every . A colimit of is an initial such cocone: for every cocone , there is a unique with for all .
Let be a final functor and let be a cocone under . For each , choose an object of the nonempty comma category and defineA morphism in the comma category gives , and the cocone identity shows that the two resulting maps are equal. Because is a connected category, a zigzag proves independence of the chosen object. The same construction applied after a morphism proves naturality, so is a cocone. Any extension must have this value because it must satisfy the cocone identity along , proving uniqueness. This is cocone extension along a final functor.
If is a colimit of , its universal cocone extends uniquely to . Restriction and extension give mutually inverse correspondences between cocones from and from , so the extended cocone is a colimit of . Therefore the existence of all colimits of shape implies the required colimits of shape .
Now suppose is a sifted category. For , the product functor is a left adjoint, so it preserves colimits. Applying this once in each variable givesThe diagonal is final, so the right side isThus preserves binary products in a category. Since is connected, the colimit of the constant singleton diagram is a singleton, so it also preserves the terminal object. It therefore preserves finite products.