# Completeness of over category (slice)

Let $$\mathcal J$$ be a (small) category (denote $$I:= \mathcal J_0$$) and $$\mathcal C$$ a category that has all (small) limits (all limits of shape $$\mathcal J$$ for all $$\mathcal J$$). Prop 3.4 states then the under category $$C\downarrow \mathcal C$$ corresponding to $$C\in\mathcal C_0$$ also has all (small) limits.

I tried doing something similar for the over category $$\mathcal C\downarrow C$$. Taking $$F:\mathcal J \to \mathcal C\downarrow C$$ and $$U:\mathcal C\downarrow C \to \mathcal C$$ the forgetful functor, we'd have by assumption that the category of $$UF$$-cones has a terminal object, call it $$(\lim UF, (k_i)_{i\in I})$$.

The hopes and dreams would be that we'd have $$(\lim UF\to C, (l_i)_{i\in I})$$ as a terminal object for the category of $$F$$-cones. I see no reason why there would be a morphism $$\lim UF\to C$$ in $$\mathcal C$$, though.

Is there another way of getting a terminal object for the $$F$$-cones?

here lay logical nonsense

• You are asking different things above and below the line. Above you want the limit in the slice to be given by the same cone as in $\mathcal C$ itself. Below you are asking for there to just be some limit in the slice. Also, why would there be an arrow from the terminal object to $C$? – Mark Kamsma Apr 12 '19 at 8:38
• @MarkKamsma eventually, the goal is to deduce that any slice has all limits, if $\mathcal C$ does. Doesn't have to be the same cone, but it was the first candidate that popped to mind. – Alvin Lepik Apr 12 '19 at 8:43
• I think that is true, at the cost of having to change the cone. I'll edit my answer in a bit. – Mark Kamsma Apr 12 '19 at 8:50

We can get a similar result to the proposition you mention, if we assume the diagram is connected and non-empty.

Proposition. Let $$I$$ be a connected and non-empty category and let $$\mathcal{C}$$ be some category that has limits of type $$I$$. Fix some object $$C$$ in $$\mathcal{C}$$. Then $$\mathcal{C}/C$$ has all limits of type $$I$$ and they are calculated in the same way as in $$\mathcal{C}$$, in the sense that the forgetful functor $$U: \mathcal{C}/C \to \mathcal{C}$$ preserves limits of type $$I$$.

Proof. Let $$F: I \to \mathcal{C}/C$$ be some diagram. Denote by $$U: \mathcal{C}/C \to \mathcal{C}$$ the forgetful functor. Then as you already noted, we have a limiting cone $$\lim UF$$ in $$\mathcal{C}$$ with projections $$p_i: \lim UF \to UF(i)$$ for each object $$i$$ in $$I$$.

Now let $$i$$ be any object in $$I$$, then $$F(i)$$ is an object in $$\mathcal{C}/C$$, so it is some arrow $$f_i: UF(i) \to C$$ in $$\mathcal{C}$$. Define $$\ell: \lim UF \to C$$ as $$\ell = f_i p_i$$. This does not depend on the choice of $$i$$, which follows from the assumption that $$I$$ is connected. (This is the point where I hoped to draw a diagram, but I cannot make it work properly. So if someone else can, please do! In the meantime, try drawing it yourself on a piece of paper.) To see this, let $$j$$ be some object in $$I$$. There is a sequence of arrows between $$UF(i)$$ and $$UF(j)$$. For every step $$k$$ in this sequence we have a projection $$p_k: \lim UF \to UF(k)$$ and an arrow $$f_k: UF(K) \to C$$, such that everything commutes and $$i$$ and $$j$$ really give the same arrow $$\ell$$.

Now we do find a good candidate for the limit in $$\mathcal{C}/C$$, namely $$\ell: \lim UF \to C$$ together with the same set of projections $$p_i$$. This does indeed form a limit. Let $$d: D \to C$$ together with projections $$q_i$$ be some cone of $$F$$ in $$\mathcal{C}/C$$. Then $$D$$ together with $$q_i$$ forms a cone in $$\mathcal{C}$$. So there is an induced morphism of cones $$u: D \to \lim UF$$. Now we only need to check that $$u$$ is indeed an arrow in $$\mathcal{C}/C$$ as well. Let $$f_i: UF(i) \to C$$ be some object in the diagram of $$F$$, then because $$q_i$$ is an arrow in $$\mathcal{C}/C$$: $$d = f_i q_i,$$ and since $$u$$ is a morphism of cones we have $$q_i = p_i u$$, so $$f_i q_i = f_i p_i u,$$ finally by the definition that $$\ell = f_i p_i$$: $$f_i p_i u = \ell u.$$ So summing up we have indeed $$d = f_i q_i = f_i p_i u = \ell u,$$ as required. QED.

If the diagram is not connected, or if it is empty, we have no hope of the above proposition being true in general. Even if we assume $$\mathcal{C}$$ to have all limits. Consider the following two examples.

Example 1. No matter what category $$\mathcal{C}$$ and object $$C$$ we start with, the category $$\mathcal{C}/C$$ always has a terminal object and it is given by $$Id_C: C \to C$$. So if $$\mathcal{C}$$ already had a terminal object $$1$$, and we take $$C$$ to be non-terminal, then the forgetful functor does not preserve the terminal object.

Example 2. Let us consider $$\mathbf{Set}$$, the category of sets. Let us consider the set $$\mathbb{N}$$ of natural numbers, together with the subsets $$E$$ and $$O$$ of even and odd numbers respectively. We can naturally find $$E$$ and $$O$$ in $$\mathbf{Set} / \mathbb{N}$$ as well, by just considering the inclusions $$E \hookrightarrow \mathbb{N}$$ and $$O \hookrightarrow \mathbb{N}$$. The product of $$E \times O$$ in $$\mathbf{Set}$$ is just their cartesian product (with the obvious projections). The product in $$\mathbf{Set} / \mathbb{N}$$ does exist, but this is the empty set (with the empty function to $$\mathbb{N}$$)! This last part will be clear in a bit, when we prove that products in $$\mathbf{Set} / \mathbb{N}$$ are given by pullbacks in $$\mathbf{Set}$$ (so in this case, by the intersection $$E \cap O$$).

If we are just interested in $$\mathcal{C}/C$$ being complete, we have the following result.

Proposition. If $$\mathcal{C}$$ is complete, then so is $$\mathcal{C}/C$$.

This result does (implicitly) appear in most books about topos theory. When proving that for any topos $$\mathcal{E}$$ the slice topos $$\mathcal{E}/X$$, by some object $$X$$ from $$\mathcal{E}$$, is again a topos, one has to show that $$\mathcal{E}/X$$ is complete (although, technically this is about being finitely complete, but it easily generalises). This part of the proof only uses completeness of $$\mathcal{E}$$. For example, a proof can be found in Sheaves in Geometry and Logic by MacLane and Moerdijk, at the start of theorem IV.7.1. I will present a (sketch of a) proof here as well, so we can link it to the proposition at the start of this answer.

Proof. As mentioned in example 1 above, the category $$\mathcal{C}/C$$ always has a terminal object. By the proposition at the start of this answer, $$\mathcal{C}/C$$ has equalisers (and they are in fact 'the same' as in $$\mathcal{C}$$). So all we need to check is products. So let $$(A_i \to C)_{i \in I}$$ be a non-empty set of objects in $$\mathcal{C}/C$$. Form their wide pullback $$P$$ in $$\mathcal{C}$$. There is only one arrow $$P \to C$$ to be considered, and this will be the desired product in $$\mathcal{C}/C$$ (check this!). We have now shown that $$\mathcal{C}/C$$ has all small products and equalisers, so it is complete. QED.

We have essentially obtained a way to calculate limits in $$\mathcal{C}/C$$. For any diagram $$F: D \to \mathcal{C}/C$$ we obtain a diagram $$F'$$ in $$\mathcal{C}$$ by just 'forgetting' that we lived in $$\mathcal{C}/C$$. So I do not mean to just apply the forgetful functor here, because we want to keep all the arrows to $$C$$ in our diagram $$F'$$ (another way to describe this would be to apply the forgetful functor, and then add all the arrows to $$C$$ back in). Now we calculate the limit $$\lim F'$$ of $$F'$$ in $$\mathcal{C}$$. Since $$C$$ was in the diagram $$F'$$, we have a projection $$\lim F' \to C$$ and this will be the limit in $$\mathcal{C}/C$$.

The connection with the propisition at the start of this answer is that if $$F$$ is non-empty connected, we do not need to keep $$C$$ in the diagram to make things work.

• I hope I understand the counter example correctly. Have you demonstrated the over category lacks a limit of type $\textbf{2}$ i.e product? – Alvin Lepik Apr 11 '19 at 17:33
• @AlvinLepik yes, that is what I intended (suggestions for clearer wording are welcome!) – Mark Kamsma Apr 11 '19 at 17:36
• The noncommuting condition allows for no equalizers to exist for the pair $af,bg: X\to C$, thus the example category $\mathcal C$ doesn't contain all limits. We assume $\mathcal C$ has limits of all shapes. – Alvin Lepik Apr 12 '19 at 7:36
• Your input is of great help, Mark. I must admit I didn't have the clearest understanding of what I was asking initially nor how simple the answer actually is. Then again, I guess, retrospectively everything is clear. – Alvin Lepik Apr 12 '19 at 14:08
• Glad to help. Also, I know the feeling. Usually when I feel like some problem was easy once its solved, I take it as a good sign: it means that a real understanding was developed along the way. – Mark Kamsma Apr 12 '19 at 14:11