The definition of the category of cones reminds me of slice categories, except that we want an arrow between cones to induce (possibly) many commutative triangles.

Can $\mathsf{Cone}(F)$ be written as a slice category?

  • 3
    $\begingroup$ The common generalisation of both is the comma category. $\endgroup$ – Zhen Lin Nov 6 '14 at 16:53
  • $\begingroup$ It seems $\mathsf{Cone}F$ is only isomorphic to a comma category though, no? $\endgroup$ – user153312 Nov 8 '14 at 10:15

Not really, but it is a subcategory of a slice category.

If $F : J → \mathscr C$ is a functor, then it is an object of the functor category $\mathscr C^J$. For every object $C$ of $\mathscr C$ there is a functor $ΔC : J → \mathscr C$, sending every object of $J$ to $C$, and every morphism to $\mathrm{id}_C$. Natural transformations from $ΔC$ to $F$ are exactly the cones from $C$ to $F$. So every object of $\operatorname{Cone} F$ is an object of the slice category $\mathscr C^J/F$.

Furthermore, $Δ$ extends to a functor $Δ : \mathscr C → \mathscr C^J$ (called the diagonal functor). If $f : C → D$ is a morphism in $\mathscr C$, then we define the natural transformation $Δf : ΔC → ΔD$ by $(Δf)_j = f$, for every object $j$ of $J$. This functor is faithful when $J$ is non-empty.

Now take two cones, say $γ : ΔC → F$ and $δ : ΔD → F$, and $f : C → D$ a morphism of cones. You can easily check that $Δf$ is then a morphism in $\mathscr C^J/F$. Since $Δ$ is faithful, you can take $\operatorname{Cone} F$ to be a subcategory of $\mathscr C/F$.

Or in other words, as Zhen Lin mentions in a comment, $\operatorname{Cone} F$ is (isomorphic to) the comma category $(Δ \downarrow F)$, and the limiting cone is the terminal object in that category.

This btw. is a useful perspective to have because it follows by certain basic results on adjoint functors that if every functor $F : J → \mathscr C$ has a limit, then there exists a functor $\operatorname{lim} : \mathscr C^J → \mathscr C$ right adjoint to $Δ$, such that $\operatorname{lim} F$ is a limit of $F$ for every functor $F : J → \mathscr C$.

| cite | improve this answer | |
  • 1
    $\begingroup$ $\Delta : \mathcal{C} \to [\mathcal{J}, \mathcal{C}]$ is not full in general. Consider, for instance, $\mathcal{J} = \{ 0, 1 \}$, so that $[\mathcal{J}, \mathcal{C}] \cong \mathcal{C} \times \mathcal{C}$. $\endgroup$ – Zhen Lin Nov 7 '14 at 8:49
  • $\begingroup$ @ZhenLin: Ugh, thanks. I had something I thought was an obvious argument on my mind, and hadn't checked how little sense it made. I've edited the answer. $\endgroup$ – user54748 Nov 7 '14 at 14:33
  • $\begingroup$ Categories for the working mathematician goes along these lines. I was just curious whether I'm missing an explicit slice category. Thanks for your answer! $\endgroup$ – user153312 Nov 8 '14 at 1:31
  • $\begingroup$ @user54748 aren't the functors in a comma category supposed to have common codomains? How is $\varDelta \downarrow F$ defined if $F$ has codomain $\mathsf{C}$ and $\varDelta$ has codomain $\mathsf {C}^\mathsf{J}$? $\endgroup$ – user153312 Apr 11 '15 at 18:44
  • 2
    $\begingroup$ @Exterior: $(Δ \downarrow F)$ is a comma category the same way the general slice category $\mathscr C/C$ is: the element is interpreted as the functor $C : 1 → \mathscr C$ from the terminal category. That in this case the element $F$ is itself a functor is coincidental, and $F$ stands for the functor $F : 1 → \mathscr C^J$. This abuse of notation can hardly lead to ambiguity here, but I should've pointed it out. $\endgroup$ – user54748 Apr 11 '15 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy