Return to Answer

Somehow this view takes over to the general case: Let $F : J \to Set$ be a functor. Regard it as a presheaf on $J^{op}$, and he map induced by $i \to j$ in $J^{op}$ as a restriction $F(j) \to F(i)$. Also call the elements of $F(i)$ sections on $i$. Then the limit of $F$ consists of compatible sections. Since I've been learning algebraic geometry, I almost always think of limits in this way.

I think this book is still one of the best introductions into category theory. I can be hard to grasp all these abstract concepts and examples, but it gets easier as soon as you get input from other areas where category theoretic ideas are omnipresent. Your example about gluing morphisms illustrates this very well.

Somehow this view takes over to the general case: Let $F : J \to Set$ be a functor. Regard it as a presheaf on $J^{op}$, and the map induced by $i \to j$ in $J^{op}$ as a restriction $F(j) \to F(i)$. Also call the elements of $F(i)$ sections on $i$. Then the limit of $F$ consists of compatible sections. Since I've been learning algebraic geometry, I almost always think of limits in this way.

I think this book is still one of the best introductions into category theory. It can be hard to grasp all these abstract concepts and examples, but it gets easier as soon as you get input from other areas where category theoretic ideas are omnipresent. Your example about gluing morphisms illustrates this very well.

This is not correct. The reason is that the index category can be rather wild and colimits in preorders don't care about that. In detail: Let $U : J \to O(X)^{op}$ be a small diagram. Then the limit is just the union $V$ of $U_j$. Thus $F$ preserves this limit iff sections on $V$ are sections on the $U_j$ which are compatible with respect to the restriction morphisms given by $U$. If $J$ is discrete and $U$ maps everything to the same open subset $V$ of $X$, then the compatible sections are $F(V)^J$, which is bigger than $F(V)$.

Finally it is important to remember that limit is just the dual concept of colimit. And often algebra and geometry appear dually at once, for example sections and open subsets in sheaves. If $(X_i,\mathcal{O}_{X_i})$ are ringed spaces and you want to find the colimit, well you can guess that you have to do: Take the colimit of the $X_i$ and the limit of the $\mathcal{O}_{X_i}$ (pullbacked to the colimit).

This is not correct. The reason is that the index category can be rather wild and colimits in preorders don't care about that. In detail: Let $U : J \to O(X)^{op}$ be a small diagram. Then the limit is just the union $V$ of $U_j$. Thus $F$ preserves this limit iff sections on $V$ are sections on the $U_j$ which are compatible with respect to the restriction morphisms given by $U$. If $J$ is discrete and $U$ maps everything to the same open subset $V$ of $X$, then the compatible sections are $F(V)^J$, which is bigger than $F(V)$.

"... I have a copy of MacLane's "Categories for the Working Mathematician," but whenever I pick it up, I can never seem to get through more than two or three pages (except in the introduction on foundations"

I think this book is still one of the best introductions into category theory. I can be hard to grasp all these abstract concepts and examples, but it gets easier as soon as you get input from other areas where category theoretic ideas are omnipresent. Your example about gluing morphisms illustrates this very well.

Assume that $X$ is a set and $X_i$ are subsets of $X$ whose union is $X$. Then it is clear how to characterize functions on $X$: These are simply functions on the $X_i$ which agree on the overlaps $X_i \cap X_j$. This can formulated in a fancy way: Let $J$ be the category whose objects are the indices $i$ and pairs of such indices $(i,j)$. It should be a preorder and we have the morphisms $(i,j) \to i, (i,j) \to j$. We have a diagram $J \to Set$, which is given by $i \mapsto X_i, (i,j) \mapsto X_i \cap X_j$. What we have remarked above says exactly that $X$ is the colimit of this diagram! In a similar fashion, open coverings can be understood as colimits in the category of topological spaces, ringed spaces or schemes. It's all about gluing morphisms.

Now what about limits? I think it is important to understand first limits in the category of sets. If $F : J \to Set$ is a small diagram, then we can consider simply the set of "compatible elements in the image" of $F$, namely $X = \{x \in \prod_j \in F(j) : \forall i \to j : x_j = F(i \to j)(x_i)\}$. A short definition would be $X = Cone(*,F)$. Observe that we have projections $X \to F(j), x \mapsto x_j$ and with these $X$ is the limit of $F$. Now the Yoneda-Lemma or just the definition of a limit tells you how you can think of a limit in a arbitrary category: $X$ is a limit of diagram $F : J \to C$ amounts to say that elements of $X$ .. erm we don't have any elements, so let's say morphisms $Y \to X$, naturally correspond to compatible elem... erm morphisms $Y \to F(i)$. In other words, for every $Y$, $X(Y)$ is the set-theoretic limit of the diagramm $F(Y)$.

the sheaf condition on a presheaf can be expressed as stating that the contravariant functor takes colimits to limits

This is not correct. The reason is that the index category can be rather wild and colimits in preorders don't care about that. In detail: Let $U : J \to O(X)^{op}$ be a small diagram. Then the limit is just the union $V$ of $U_j$. Thus $F$ preserves this limit iff sections on $V$ are sections on the $U_j$ which are compatible with respect to the restriction morphisms given by $U$. If $J$ consists of two objects $i,j$ with two morphisms $i \to j$ (and the identities of $i,j$), then these compatible sections are just two sections on $U_2$, thus not sections on $V=U_2$.

Assume that $X$ is a set and $X_i$ are subsets of $X$ whose union is $X$. Then it is clear how to characterize functions on $X$: These are simply functions on the $X_i$ which agree on the overlaps $X_i \cap X_j$. This can be formulated in a fancy way: Let $J$ be the category whose objects are the indices $i$ and pairs of such indices $(i,j)$. It should be a preorder and we have the morphisms $(i,j) \to i, (i,j) \to j$. Consider the diagram $J \to Set$, which is given by $i \mapsto X_i, (i,j) \mapsto X_i \cap X_j$. What we have remarked above says exactly that $X$ is the colimit of this diagram! In a similar fashion, open coverings can be understood as colimits in the category of topological spaces, ringed spaces or schemes. It's all about gluing morphisms.

Now what about limits? I think it is important first to understand limits in the category of sets. If $F : J \to Set$ is a small diagram, then we can consider simply the set of "compatible elements in the image" of $F$, namely

$X = \{x \in \prod_j F(j) : \forall i \to j : x_j = F(i \to j)(x_i)\}$.

A short definition would be $X = Cone(*,F)$. Observe that we have projections $X \to F(j), x \mapsto x_j$ and with these $X$ is the limit of $F$. Now the Yoneda-Lemma or just the definition of a limit tells you how you can think of a limit in an arbitrary category: That $X$ is a limit of a diagram $F : J \to C$ amounts to say that elements of $X$ .. erm we don't have any elements, so let's say morphisms $Y \to X$, naturally correspond to compatible elem... erm morphisms $Y \to F(i)$. In other words, for every $Y$, $X(Y)$ is the set-theoretic limit of the diagramm $F(Y)$. I hope that this makes clear that the concept of limits in arbitrary categories is already visible in the category of sets.

"...the sheaf condition on a presheaf can be expressed as stating that the contravariant functor takes colimits to limits"

This is not correct. The reason is that the index category can be rather wild and colimits in preorders don't care about that. In detail: Let $U : J \to O(X)^{op}$ be a small diagram. Then the limit is just the union $V$ of $U_j$. Thus $F$ preserves this limit iff sections on $V$ are sections on the $U_j$ which are compatible with respect to the restriction morphisms given by $U$. If $J$ is discrete and $U$ maps everything to the same open subset $V$ of $X$, then the compatible sections are $F(V)^J$, which is bigger than $F(V)$.