Cocomplete base category implies cocomplete slice category

I'm having trouble with the final step of the proof that if $\mathsf C$ is a cocomplete category, so is each of its slice categories.

Here's the proof given in Borceux's Handbook of Categorical Algebra:

I understand everything except the very last sentence. Why is the conclusion immediate?

• Try to prove the universal property for the alleged colimit of $F$ in $\mathscr{C}/I$. – Marco Vergura Jun 2 '15 at 9:38
• There is a typo, it should be $\gamma_D : G(D) \to I$. – Martin Brandenburg Jun 3 '15 at 6:44
• It is not immediate that $((L,\lambda),(s_D))$ is a colimit of $F$. It is easy to see, but one really has to check this. Details can be found for example in Mac Lane's book. But this is also a good exercise which one can easily solve. – Martin Brandenburg Jun 3 '15 at 6:45

By construction $$\left( (L,\lambda),(s_D)_{D \in \mathscr{D}} \right)$$ is a cocone on $$F$$. If $$\left( (M,m),(q_D)_{D \in \mathscr{D}} \right)$$ is another cocone on $$F$$, then $$\left( M,(q_D)_{D \in \mathscr{D}} \right)$$ is a cocone on $$G$$. By the universal property of colimits there is a unique factorization $$l : L \to M$$, such that $$l \circ s_D = q_D$$ for all $$D \in \mathscr{D}$$. It is now left to prove that $$\lambda = m \circ l$$. We have $$m \circ l \circ s_D = m \circ q_D = \gamma_D = \lambda \circ s_D.$$ The result now follows from the uniqueness of factorization $$L \to I$$ (It is the dualized version of proposition 2.6.4 in Borceux's book). This reasoning is fairly easy and could be found in earlier proofs.
Typo: As it was mentioned in the comments, morphisms $$\gamma_D$$ must map $$GD \to I$$.