Lemma 4.14.8 (002K)—The Stacks project

Lemma 4.14.8. Suppose that $M : \mathcal{I} \to \mathcal{C}$, and $N : \mathcal{J} \to \mathcal{C}$ are diagrams whose colimits exist. Suppose $H : \mathcal{I} \to \mathcal{J}$ is a functor, and suppose $t : M \to N \circ H$ is a transformation of functors. Then there is a unique morphism

\[ \theta : \mathop{\mathrm{colim}}\nolimits _\mathcal {I} M \longrightarrow \mathop{\mathrm{colim}}\nolimits _\mathcal {J} N \]

such that all the diagrams

\[ \xymatrix{ M_ i \ar[d]_{t_ i} \ar[r] & \mathop{\mathrm{colim}}\nolimits _\mathcal {I} M \ar[d]^{\theta } \\ N_{H(i)} \ar[r] & \mathop{\mathrm{colim}}\nolimits _\mathcal {J} N } \]

commute.

Comments (4)

Comment #76 by Keenan Kidwell on October 15, 2012 at 23:58

I think the $t_{H(i)}$ in the diagram should be $t_i$ .

Comment #83 by Johan on October 16, 2012 at 16:55

Well, I am following the notation: If $t : F \to G$ is a transformation of functors then $t$ is made up of maps $t_X : F(X) \to G(X)$ where $X$ varies over objects of the source category of $F$ and $G$ . Since in the statement of the lemma $M$ and $N \circ H$ have source $\mathcal{J}$ , I think the notation makes sense. I am not saying the notation is perfect though...

Comment #88 by Keenan Kidwell on October 16, 2012 at 17:11

Both $M$ and $N\circ H$ have source category $I$ , right? So wouldn't writing $t_i$ be in accordance with the notation?

Comment #89 by Johan on October 16, 2012 at 17:45

Oops! Yes. Fixed. Thanks!

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7 comment(s) on Section 4.14: Limits and colimits

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