Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 119 / 2 / i / Solution

Let ${\lambda }_A:GF(A)\to Z$ be the given cocone under a diagram. For $B\in \mathcal{D}$, choose an object $(A,u:B\to FA)$ of the comma category $(B\downarrow F)$ and defineSuch a choice exists because $F$ is a final functor. If $a:A\to A^{\prime }$ gives a morphism $(A,u)\to (A^{\prime },u^{\prime })$, so $F(a)u=u^{\prime }$, the cocone relation givesThe connected category condition makes the result independent of the choice along any zigzag.

For $v:B\to B^{\prime }$, choose $(A,u:B^{\prime }\to FA)$; then $(A,uv)$ is available for $B$, giving ${\bar{\lambda }}_B={\bar{\lambda }}_{B^{\prime }}G(v)$. Thus the maps form a cocone under a diagram on $G$. Choosing $(A,1_{FA})$ shows that restriction recovers ${\lambda }_A$. Any extension must satisfy ${\bar{\lambda }}_B={\bar{\lambda }}_{FA}G(u)={\lambda }_{A}G(u)$, proving uniqueness. This is cocone extension along a final functor.

If $GF$ has a colimit, extend its universal cocone in this way. Every cocone on $G$ restricts to one on $GF$, factors uniquely through that colimit, and then agrees with its extension by uniqueness. Hence the same vertex is a colimit of $G$. In particular,and existence of all $\mathcal{C}$-shaped colimits in $\mathcal{E}$ implies existence of all $\mathcal{D}$-shaped colimits.