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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:

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

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    $\begingroup$ Try to prove the universal property for the alleged colimit of $F$ in $\mathscr{C}/I$. $\endgroup$ – Marco Vergura Jun 2 '15 at 9:38
  • $\begingroup$ There is a typo, it should be $\gamma_D : G(D) \to I$. $\endgroup$ – Martin Brandenburg Jun 3 '15 at 6:44
  • $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Jun 3 '15 at 6:45
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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 $GD \to I$.

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