Let $D$ be a limit-complete category. My vague question is: given two diagrams in $D$, what comparisons between them induce a morphism of their limits? I'm especially interested in the case that the comparison has something to do with colimits.

Here is a well-known case in which the comparison has nothing to do with colimits. Let $I,J$ be small categories, and let $X:I\to D$ and $Y:J\to D$ be functors. Given a functor $f:J\to I$ and a natural transformation $\alpha:X\circ f\to Y$, there is an induced morphism between their limits $\lim(f,\alpha):\lim(X)\to \lim(Y)$

But today I came across another case in which one gets such a map between limits. Let $D={\bf Set}$ be the category of sets. Suppose $I$ is the W-shaped category $M\to N\leftarrow P\to Q\leftarrow R$ in $D$. Suppose $J$ is the cospan $M\to S\leftarrow R$, and let $g:I\to J$ be the functor sending $M\mapsto M, R\mapsto R$ and $N,P,Q\mapsto S$. \begin{align} I:\hspace{.5in}M\to \fbox{$N\leftarrow P\to Q$}\leftarrow R\\\\ g\downarrow\hspace{.8in}\\\\ J:\hspace{.5in}M\xrightarrow{\hspace{.5in}} S\xleftarrow{\hspace{.5in}} R \end{align} Given a functor $X:I\to D$, let $Y=Lan_gX:J\to D$ be the left Kan extension of $X$ along $g$. To my (perhaps very naive) surprise, I get a morphism of limits in this case too, $\lim(X)\to \lim(Y)$. To be explicit, this map sends a tuple $(m,p,r)\in \lim(X)$ to the pair $(m,r)\in \lim(Y)$.

Question 1 (Sets): Is it true for any small categories $I, J$, and $X:I\to{\bf Set}$ that a functor $g:I\to J$ induces a function $\lim(X)\to \lim(Lan_gX)$?

Question 2 (General): Is it true for general $D$ that for any categories $I, J$, and $X:I\to D$ that a functor $g:I\to J$ induces a morphism $\lim(X)\to \lim(Lan_gX)$?

Question 3 (Vague): Is this part of any more general result about the interaction between limits and colimits?

Any references would be appreciated.


The existence of that sort of morphism between limits relates to some properties of the functor $g : I \rightarrow J$.

Given functors $h: I \rightarrow D$, $l: J \rightarrow D$ and a natural transformation $t : h \rightarrow lg$ we want to construct a morphism $Lim(h) \rightarrow Lim(l)$. Let $z = Lim(h)$, and let $s : z \rightarrow h$ be the universal cone. Suppose that for each object $x$ of $J$ a choice of an object $g^{-1}(x)$ in its fiber over $g$ is made. Then, for any $x$ we can define a morphism $z \rightarrow l(x)$ by $$z \xrightarrow{s_{g^{-1}(x)}} h(g^{-1}(x)) \xrightarrow{t_{g^{-1}(x)}} lg(g^{-1}(x)) = l(x)$$ If this family is a cone $z \rightarrow l$, then we get a morphism $z \rightarrow Lim(l)$ that we want. This family is a cone for example when for each arrow $f :x \rightarrow x'$ in $J$ there is a chain of arrows in $I$ $$g^{-1}(x) \xrightarrow{k_1} \xleftarrow{k'_1}\xrightarrow{k_2}...\xrightarrow{k_{n}}\xleftarrow{k'_n} g^{-1}(x')$$

such that $g(k'_i) = 1$ and $g(k_n)...g(k_2)g(k_1) = f$. For then we can establish $l(f)s_{g^{-1}(x)}t_{g^{-1}(x)} = s_{g^{-1}(x)}t_{g^{-1}(x)}$. For example for $n = 1$, omitting indexes for $s$ and $t$, this is shown by $$l(f)ts = lg(k_1)ts = th(k_1)s = ts = th(k'_1)s = lg(k'_1)ts = ts.$$

In your example you take $g^{-1}(M) = M$, $g^{-1}(R) = R$, and for $g^{-1}(S)$ you can take either $N$, $P$ or $Q$. Pretty much this is why the morphism between the limits exists.


I'd say the answer to Q1 is no in general. Take $I=\{1\}$ to be a discrete category on one object, $J=\{0\rightarrow 1\}$, and $g\colon I\rightarrow J$ the inclusion. Let $X\colon I\rightarrow \mathbf{Set}$ be a non-empty set. I think that $\operatorname{Lan}_gX$ is the inclusion of the empty subset $\varnothing\rightarrow X$ and there is no map $X=\lim X\rightarrow \varnothing=\lim(\varnothing\rightarrow X)$.

  • $\begingroup$ Thanks, that's helpful! But then is there any general situation that encompasses the phenomenon described above? For example, I think my functor $g:I\to J$ is initial, whereas yours is not. Could that be what makes it work? $\endgroup$ – David Spivak Apr 26 '14 at 21:27

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.