# In the absence of limits from the codomain, do evaluation functors still preserve limits?

Consider a diagram $K:\mathsf J\longrightarrow [\mathsf C,\mathsf D]$ as well as the evaluation functors $\operatorname{ev}_C:[\mathsf C,\mathsf D]\to \mathsf D$. The fact limits in functor categories are computed pointwise amounts to the following.

Assuming each $\operatorname{ev}_C\circ K$ has a limit, these limits come together to define a limit of $K$. Consequently each evaluation functor preserves limits and the evaluation functors "jointly create limits".

As an exercise I've been told to prove that without any assumptions, the evaluation functors preserve limits (and colimits). However, I just don't see why this should be true. The existence of a limit for each composite $\operatorname{ev}_C\circ K$ is used to construct to limit upstairs (using universal properties), and that is what makes preservation obvious. I don't understand why the limit of the functor $K$ should evaluate to limits of $\operatorname{ev}_C\circ K$ in general.

Am I wrong or is there a counterexample? (I started fiddling with diagrams, but I'm hoping for some trivial counterexample.)

• Start by asking yourself when is it automatic that the evaluation functor preserves limits. What condition on $D$ is required? Work from there. Commented Apr 19, 2018 at 14:01
• @IttayWeiss I'm not sure I understand your advice. Certainly if $\mathsf D$ is complete then the limits of the composites $\operatorname{ev}_C\circ K$ exist and we may proceed with the usual proof. Alternatively, if $\mathsf D$ is complete may consider left and right adjoints to $\operatorname{ev}_C$ given by pointwise left and right Kan extensions along the points of $\mathsf C$. How does this help if we drop assumptions on $\mathsf D$ (that is what I'm asking)? Commented Apr 19, 2018 at 14:44
• Try to directly write down the right adjoint to evaluation. What minimal condition guarantees its existence? Commented Apr 19, 2018 at 16:31
• @IttayWeiss I don't know how to write down the right adjoint. I have no suspect for what it might be concretely, especially without any limits with which to produce a weighted limit formula for the Kan extension. Could you please elaborate? Commented Apr 19, 2018 at 20:09

Take $\mathbf C$ to be the category with one non-identity morphism, $\mathbf D$ to be the category with three non-identity morphisms $A\to B\leftleftarrows C$, and $\mathbf J\xrightarrow{K}[\mathbf C,\mathbf D]=\mathbf D^\to$ to be two copies of the morphism $A\to B$ ($\mathbf J$ is the category with two objects).

Then as a morphism $A\to B$ is its own square because the only morpisms to it are commutative squares from itself and $A\xrightarrow{\mathrm id_A}A$, but its codomain $B$ does not have a square because the pair of morphisms $B\leftleftarrows C$ only factor jointly through themselves, while the redundant pair of morphisms $A\to B$ only factor through $A$ and $B$.

Below is my reasoning for arriving at this minimal counter-example.

In the absence of limit conditions on a category, a good replacement for the notion of a limit-preserving functor is the notion of a flat functor. Explicitly, given a diagram $\mathbf J\xrightarrow{K}\mathbf E$, a functor $\mathbf E\xrightarrow{F}\mathbf D$ is $K$-flat if any cone over the diagram $\mathbf J\xrightarrow{K}\mathbf E\xrightarrow{F}\mathbf D$ factors through the image of a cone over $\mathbf J\xrightarrow{K}\mathbf E$. This notion is good because a) if $\mathbf J\xrightarrow{K}\mathbf E$ has a (weak) limit, then $\mathbf E\xrightarrow{F}\mathbf D$ is $K$-flat if and only if it preserves this (weak) limit, b) right adjoints are flat for all diagrams, c) the general adjoint functor theorem says that when $\mathbf E$ is Cauchy-complete locally small, then $\mathbf E\xrightarrow{F}\mathbf D$ has a left adjoint if and only if $\mathbf E\xrightarrow{F}\mathbf D$ is flat for small diagrams and objects of $\mathbf D$ satisfy the solution set condition (i.e. have "small prereflections") with respect to $\mathbf E\xrightarrow{F}\mathbf D$.

In your situation, a cone over $\mathbf J\xrightarrow{K}[\mathbf C,\mathbf D]\xrightarrow{\mathrm{ev}_{c_0}}\mathbf D$ is of course an object $d\in\mathbf D$ equipped with a family of morphisms $d\to K(j,c_0)$ natural in $j$.

On the other hand, a cone over $\mathbf J\xrightarrow{K}[\mathbf C,\mathbf D]$ is a functor $\mathbf C\xrightarrow{X}\mathbf D$ equipped with natural transformations $X\Rightarrow K$, i.e. a family of morphisms $X(c)\to K(j,c)$ in $\mathbf D$ natural in both $j$ and $c$.

Thus, evaluation at $c_0$ is flat if any family of morphisms $d\to K(j,c_0)$, natural in $j$, factors as $d\to X(c_0)\to K(j,c_0)$ for a family of morphisms $X(c)\to K(j,c)$ natural in $j$ and $c$. For a minimal counterexample, we see that $J$ being empty or the category with one object cannot work, hence we should take it to be at least the category with two objects. Then we are reduced to ensuring that a pair of morphisms $d\rightrightarrows K(j_0,c_0), K(j_1,c_0)$ do not jointly factor as $d\to X(j,c_0)\to K(j,c_0)$ for a family of morphisms $X(j,c)\to K(j,c)$ that is natural in $c$.

Because the only reason this would break now is naturality, for a minimal counter-example we can take $\mathbf C$ to be the category with two objects and a single non-identity morphism between them. Then the diagram $K$ is simply a pair of morphisms in $\mathbf D$, and we want to set up $\mathbf D$ so that the two morphisms have a product but either the domain or codomain of that product is not the product of their domains or codomains.

Keeping in mind that the two morphisms don't have to be distinct, I arrived at the above counter-example.

• Re : "set up $D$ so that the two morphisms have a product but either the domain or codomain of that product is not the product of their domains or codomains", it's perhaps worth pointing out that this can only work for the codomains : the "domain" functor from the category of arrows to $D$ does in fact preserves limits, because it is the right adjoint of the functor from $D$ to its category of arrows that map every object to its identity arrow. Commented Apr 20, 2018 at 12:16

I think the following provides a counterexample : take $C$ to be the poset $\{0\leq 1\}$, and $D$ be the poset $\{x_1,x_2,x_3,y_1,y_2,y_3,z\}$, where $x_i\leq y_i$ for all $i$, $x_1\leq x_2$, $x_1\leq x_3$ (similarly for the $y_i$), and moreover $z\leq y_2$ and $z\leq y_3$. The functor category $[C,D]$ is the same thing as the category of arrows of $D$, which, since $D$ is a poset, is the same thing as pairs $(a,b)$ for which $a\leq b$, and there is a (unique) arrow $(a,b)\to (a',b')$ if and only if $a\leq a'$ and $b\leq b'$.

Now $(x_1,y_1)$ is the product of $(x_2,y_2)$ and $(x_3,y_3)$; indeed, if $(a,b)$ has arrows to $(x_2,y_2)$ and $(x_3,y_3)$, then we must have $a\leq x_2$ and $a\leq x_3$, which forces $a=x_1$, and since $x_1\leq b\leq y_2$ and $b\leq y_3$, we must have either $b=x_1$ or $b=y_1$. Thus in any case, we have a (necessarily unique) arrow $(a,b)\to (x_1,y_1)$. But note that $y_1$ is not actually the product of $y_2$ and $y_3$, since there is no arrow between $y_1$ and $z$.

• Dear Arnaud, thank you very much for this answer. I have accepted the answer by Vladimir Sotirov because I feel it is exceptionally sharp and instructional. Commented Apr 19, 2018 at 23:23