Exercise 5.3.12 in Leinster asks to prove this result:

Let $F:\mathscr A\to\mathscr B$ be a functor and $I$ a small category. Suppose that $\mathscr B$ has, and $F$ creates, limits of shape $I$. Then $\mathscr A$ has, and $F$ preserves, limits of shape $I$.

By "creation of limits" I mean "strict creation of limits". Leinster just calls it "creation of limits". (See Remark 5.3.7.)

Suppose $D:I\to\mathscr A$ be a diagram.

I think I understand how the existence of limits of shape $I$ in $\mathscr A$ follows -- we know that the limit cone $(\lim F\circ D\to FD(i))_{i\in I}$ exists in $\mathscr B$, and since $F$ creates limits, it follows that the "preimage" of this limit cone under $F$ is a limit cone $(\lim D\to D(I))_{i\in I}$ in $\mathscr A$.

But I'm not sure if this proof of the fact that that $F$ preserves limits of shape $I$ is right. Suppose $(A\to D(I))_{i\in I}$ is a limit cone on $D$ in $\mathscr A$. I need to show that its "image", $(F(A)\to FD(i))_{i\in I}$, is a limit cone of $F\circ D$. It's known that $F\circ D$ has a limit cone $(\lim F\circ D\to FD(i))_{i\in I}$ and that it is the "image" of a unique limit cone a limit cone on $D$. Since $(A\to D(I))_{i\in I}$ is itself a limit cone on $D$, the image of this cone must be equal to $(\lim F\circ D\to FD(i))_{i\in I}$. But on the other hand, this image is equal to $(F(A)\to FD(i))_{i\in I}$. Thus $(F(A)\to FD(i))_{i\in I}$ is a limit cone.

Does this look right? In the last paragraph I interpreted uniqueness in the definition of (strict) creation of limits in terms of strict equality of cones. I'm not sure if it's okay even under the definition of strict creation of limits discussed in Remark 5.3.7.


1 Answer 1


No, your argument doesn't convince me. Indeed there is a unique cone in the domain which gets mapped to $(\lim F\circ D\to FD(i))_{i\in I}$ under $F$, but nothing says that this cone must be the limit cone $(A\to D(I))_{i\in I}$ you started with. In principle there is more than one limit cone in the domain. You are missing a small step.

Let me first simplify the notation. It helps to think of a cone with apex $a$ over the functor $D$ as of a natural transformation $a \Rightarrow D$. Here $a$ denotes the constant functor $I\to \mathscr A$ which sends every morphism to $1_a$. You start with a limit cone $\lambda: l\Rightarrow D$ in $\mathscr A$. You want to show that its image $F\lambda:Fl \Rightarrow FD$ is also limit cone. As you have noticed, by assumption $FD$ has a limit in $\mathscr B$. Denote it by $\mu: b \Rightarrow FD$. Now $F$ strictly creates limits, which means there is one and precisely one cone $\nu: a \Rightarrow D$ upstairs which maps to $\mu$ under $F$, and this cone is a limit cone. But now you have two limits of $D$ in $\mathscr A$. This means there must be a unique isomorphism $h: a\to l$ which is compatible with the structure maps of the two limits. $Fh$ is an isomophism $Fl \to b$ downstairs and it is compatible with $F\lambda$ and $\mu$ respectively. You already know that $\mu: b \Rightarrow FD$ is a limit, so it follows that $F\lambda : Fl \Rightarrow FD$ is a limit too, which is what you wanted to show.

  • $\begingroup$ I caught myself not understanding what an isomorphism of limit cones (or just of cones) is. If cones are natural transformations, then isomorphisms between them should be morphisms in the category of natural transformations, but I don't think Leinster's reader is supposed to know/use them. $\endgroup$
    – user557
    May 18, 2021 at 22:48
  • $\begingroup$ (Of course there is only one reasonable definition of isomorphism of cones, but I don't see why this definition fits into the general theory. I.e. I don't see what it is a particular case of.) $\endgroup$
    – user557
    May 18, 2021 at 22:56
  • $\begingroup$ This has to do with representable functors. The category of cones over $D$ in $\mathcal A$ is the category of elements of the functor $\mathcal A^I(\Delta -, D)$. Would it help you if I edit my answer to explain this? $\endgroup$
    – Nico
    May 19, 2021 at 7:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.