Limit in the category of sets In Vakil's notes on algebraic geometry, to define inverse limit, he started with a functor $F:C\to S$ where $C$ is a small category and $S$ is any category (small means objects and morphisms are sets). Later there is an exercise which asks us to prove that in the category of sets the following is the inverse limit $$\left\\{(a_i )_{i∈I} ∈ \prod_{i\in I} A_i: F(m)(a_j ) = a_k \text{ for all }m ∈ Mor_I (j,k) \in Mor(I)\right\\}.$$ Now my question is 1) Why is the condition of small category on the indexing set and not on the category $S$? 2) What does he mean by $F(m)(a_j ) = a_k $ for all $m\in Mor_I (j,k) \in Mor(I)$. (shouldn't it be : $ F(m)(A_j ) = A_k $ for all $m\in Mor_I (j,k) \in Mor(I)$ as the functor maps the indexing category to sets and $a_i$ is just an element of the set $A_i$?)

(1) Well, why would you want to require $S$ to be small? It is possible to define inverse limits when both $C$ and $S$ are large, if you like. However, usually if you want to be sure that inverse limits actually exist, you need to assume $C$ is small. For instance, if $S$ is sets, then the formula only makes sense when $C$ is small, since you need the index set $I$ of the product to be a set to be sure the product itself is a set.

(2) Here $a_j$ is an element of $A_j$ for each $j$. Since $m$ is a morphism $j\to k$, $F(m)$ is a morphism $A_j\to A_k$, so $F(m)(a_j)$ is an element of $A_k$. So it makes sense to ask whether $F(m)(a_j)$ and $a_k$ are equal, since they are both elements of $A_k$. A tuple $(a_i)_{i\in I}$ is only an element of the inverse limit set if this condition is true for all $j,k$ and all morphisms $m:j\to k$.