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I read, that a comma category $\Delta \downarrow D$ is a category, where the cones are just morphisms. $\Delta$ is the diagonal functor to category of diagrams, $D$ is the diagram.

However, a requirement for a comma category $S \downarrow T$ is that, $S, T$ have a matching codomain, however:

$$ \Delta: \mathcal{C} \to \mathcal{C}^\mathcal{J} \\ D: \mathcal{J} \to \mathcal{C} $$

What am I missing?

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1 Answer 1

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Here $D$ is not being thought of as a functor $\mathcal{J}\to\mathcal{C}$. Instead, it is being considered a functor $1\to\mathcal{C}^\mathcal{J}$, where $1$ is the category with one object and one morphism. Such a functor is of course essentially the same thing as an object of $\mathcal{C}^\mathcal{J}$, so any diagram canonically determines such a functor.

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  • $\begingroup$ Is $D(\ast)$ then what previously was $D$? $\endgroup$
    – hgiesel
    Nov 30, 2017 at 23:34
  • $\begingroup$ Yes, that's correct. $\endgroup$ Nov 30, 2017 at 23:35

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