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I am relatively new to category theory and was wondering about the following problem:

Can I consider a limit as an initial object in some categories?

Let $\mathscr{C}$ be a category and $\mathbf{J}$ a small index category. Let $T \colon \mathbf{J} \to \mathscr{C}$ be a diagram of shape $\mathbf{J}$, $\Delta_{\mathbf{J}} \colon \mathscr{C} \to \mathscr{C}^{\mathbf{J}}$ the diagonal functor and $\mathrm{Hom}_{\mathscr{C}^{\mathbf{J}}}(\square,T)\colon \mathscr{C}^{\mathbf{J}} \to \mathbf{Set}$ the contravariant Hom-functor. I know that usually, the image of a functor is not a category, but in this case, it seems to me that the image of the diagonal functor is one and I suppose the same holds for the Hom-functor (but I am not sure about that). If that is the case, I could consider the category $\mathrm{Hom}_{\mathscr{C}^{\mathbf{J}}}(\Delta_\mathbf{J}(\mathscr{C}),T)$, which would then correspond to the category of cones on $T$. In that case, wouldn't an initial object in that category correspond to a limit of the diagram $T$? Am I right or am I may be missing out on some categorical subtleties?

Sylv_
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  • Yes, you're right, I don't know what I was thinking there... – Arnaud D. Oct 07 '20 at 12:11
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    A limit cone is a terminal object in the category of cones over the given diagram. – Zhen Lin Oct 07 '20 at 15:47
  • @ZhenLin As I see it, suppose you take an initial object in the category of cones over a diagram, which then should correspond to an object of the form $\mathrm{Hom}{\mathscr{C}^{\mathbf{J}}}(\Delta{\mathbf{J}}(L),T)$ (If I am correct) and then take a cone $\mathrm{Hom}{\mathscr{C}^{\mathbf{J}}}(\Delta{\mathbf{J}}(C),T)$, then since the Hom-functor is contravariant, the unique map from the inital object to that cone is given by a morphism of the form $\mathrm{Hom}{\mathscr{C}^{\mathbf{J}}}(\Delta{\mathbf{J}}(f),T)$, with a unique $f \colon C \to L$. Why a terminal object? – Sylv_ Oct 07 '20 at 21:34
  • Contravariance of the hom-functor is irrelevant. Your construction of the category of cones is wrong. It is the comma category $(\Delta \downarrow T)$. – Zhen Lin Oct 07 '20 at 22:25
  • @ZhenLin If I get what you are telling me, your functor $\Delta$ is the composition of my $\Delta_\mathbf{J}$ with the codomain functor, i.e. a category discretisation functor? Could it simply be that our categories are dually-equivalent? – Sylv_ Oct 08 '20 at 10:09
  • By definition, functors must preserve identities and composition; therefore, the image is always a category. Otherwise, it wouldn't be useful to define a diagram as a functor. – Kevin P. Barry Oct 24 '20 at 13:18
  • @KevinP.Barry This is not true, morphisms which were not composable in the inital category can become composable in the image. Under certain conditions, the image is a category, but this is linked to my question: is the image of the functor defined above a category? – Sylv_ Oct 26 '20 at 15:15
  • If such morphisms "become" composable in the image, that composition must exist in the codomain, because by definition the codomain of a functor is a category. So, if you want it to be a category then it can be; otherwise, there isn't much point using a functor. – Kevin P. Barry Oct 27 '20 at 16:16
  • @KevinP.Barry You say that "by definition" the image of a functor is a category, but this is not the case, see link . – Sylv_ Oct 29 '20 at 08:18
  • I said that by definition the codomain is a category, which isn't the same as it's image. The point being that you should always be able to "complete" the image. – Kevin P. Barry Oct 30 '20 at 10:31

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