What does it mean that a functor preserves infinite limits?

Question

What does it mean that a functor preservers infinite limits? Can you please give an example of a functor which preserves finite limits but not infinite ones?

Answer 1

In general, a functor $F:\def\sC{\mathscr{C}}\sC\to\def\sD{\mathscr{D}}\sD$ is said to preserve the limit $\varprojlim A$ of some diagram (functor) $A:\def\sJ{\mathscr{J}}\sJ\to\sC$ if $F(\varprojlim A)$ is a limit for the diagram $F\circ A:\sJ\to\sD$. Put differently, $F$ preserves the limit $\varprojlim A$ if $\varprojlim F\circ A$ exists in $\sD$ and the canonical morphism $$ F(\varprojlim A) \to \varprojlim F\circ A $$ is an isomorphism.

If $F$ preserves the limit of all diagrams $A:\sJ\to\sC$ where $\sJ$ is finite (i.e., has finitely many objects and morphisms), then $F$ is said to preserve finite limits. If $F$ preserves all diagrams $A:\sJ\to\sC$ where $\sJ$ can be infinite (but possibly still smaller than some infinite cardinal $\kappa$), then you can say $F$ preserves infinite (or more precisely, $\kappa$-small) limits.

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Here is a simple example of a functor $F:\sC\to\sD$ that preserves finite but not infinite limits. Let $\sC$ be the category whose objects are the elements of $\def\N{\mathbb{N}}\N\cup\{\infty\}$, and there is a unique morphism $n\to m$ iff $n\geq m$. Then, the limit of any functor $A:\sJ\to\sC$ is just going to be given by $\varprojlim A = \sup_{j\in\sJ}A(j)$. In particular, finite limits are just computed by taking the maximum of all the elements involved, and the terminal object is $0$.

Now, let $\sD$ be the category obtained by taking $\sC$ and adding another point $\hat\infty$ which is bigger than everything else; that is, there is a unique morphism $\hat\infty\to n$ for every $0\leq n\leq\infty$. Similarly, limits in $\sD$ are computed by taking suprema.

Take $F:\sC\to\sD$ to be the functor where $F(n)=n$ for finite $n$, and $F(\infty)=\hat\infty$. You can check that this functor preserves finite limits (this is obvious if $\varprojlim A$ is a finite integer, and $\varprojlim A=\infty$ only if $A(j)=\infty$ for some $j$).

However, it does not preserve infinite limits. In particular, take $\sJ$ to be the full subcategory of $\sC$ spanned by the natural numbers, and let $A:\sJ\to\sC$ be the canonical inclusion. Then, $\varprojlim A = \sup_{n\in\N}n = \infty$, and therefore $F(\varprojlim A)=\hat\infty$.

However, $\varprojlim F\circ A = \sup_{n\in\N}n=\infty<\hat\infty$ in $\sD$, meaning that the canonical morphism $F(\varprojlim A)\to\varprojlim F\circ A$ is the morphism $\hat\infty\to\infty$, which is not an isomorphism.