Intuition on Algebraic geometry and categories theory This semester I do a course in advanced algebraic geometry (chapters 2-5 of hartshorne ) and for now is very hard for me to do exercises (each week in the course we need to do 14 exercises of hartshorne) for example it's get me 4 hours to do 2 exercises I think the reason why it is very difficult for me to do exercises is because I have no intuition on the following subjects: : direct limit , stalk and sheafification I will be very grateful if someone could help me to better understand these subjects and to have a good intuition for these subjects.
Thanks in advance ! 
 A: The idea of direct limit is to define an object $X$ by describing the morphisms from this objet to all the other objets of the category. For example, the coproduct of $A$ and $B$ is a direct limit, which is defined by saying that a morphism from $A \amalg B$ to $Y$ is defined by a morphism from $A$ to $Y$ and another morphism from $B$ to $Y$. This gives the notion of disjoint union for sets, or topological spaces, free products for groups, directs sum for modules,... 
In some way, you can picture the genral colimit by a coproduct with some identifications, or relations (for example, the amalgameted products for groups, or the tensor product for rings)   
The stalk of a sheaf at a point $x$ is another example which is easily understood with the sheaf $\mathcal{F}$ of continuous functions(to any other space): here, you are only interseted in the local behavior of functions near a point, so you consider all the functions defined on a neighbourhood of $x$, and you to want identify two functions which coincide on some (smaller) neighbourhood of $x$ (thus the name "germ" for the elements of the stalk). The colimit is the appropriate notion to define the stalk, since for any open neighbourhood $U$ of $x$, and $V\subset U$, there is a restriction morphism $\mathcal{F}(U) \rightarrow \mathcal{F}(V)$. From the informal description, we see that any function of $\mathcal{F}(U)$ should be identified with its image in $\mathcal{F}(V)$ in the stalk $\mathcal{F}_x$. 
So we naturally define the category $O_x$ of the open sets of $X$ containing $x$, and we define the stalk $\mathcal{F}_x$ as the colimit of the diagram obtained via the restriction of $\mathcal{F}$ to $O_x$.
For the sheafification, you can take the following comparison: let $M$ be the category of metric spaces, and $C$ be the category of complete metric spaces. Obvisoulsy we have an embedding ("forgetfull functor") $C\rightarrow M$, and on the other hand, there is an operation called completion, which gives a functor $M\rightarrow C$. The important fact is that any map from some metric space to a complete metric space factorizes uniquely through the completion of the first space. Somehow, the completion $\bar{X}$ of $X$ is the complete space closest to $X$, and the functors $C\rightarrow SET$ given by $Hom_M(X,-)$ and $Hom_C(\bar{X},-)$ are naturally isomorphic (we say that $Hom(X,-)$ is represented by $\bar{X}$). In categorical words, with have defined an adjoint pair of functors, the completion being left adjoint to the forgetfull functor. What is intersting about them, is that left adjoints preserve colimits, so you can compute colimits of complete metric spaces as colimits of metric spaces, and then complete the resulting space to get the colimit as complete metric space.
This is the same idea with preseaves and sheaves: sheaves are presheaves, and you can seafify presheaves into sheaves. The point is that presheaves are easier to define (nothing to check), and the colimits are easier to compute in the category of presheaves than in the category of sheaves. The fact that the sheafification is a left adjoint functor tells that you can compute colimits in the category of presheaves, and then sheafify, which is often easier, and gives a convenient way to talk about algebraic operations defined by universal properties (tensor products, reduction, quotient,...) on sheafs.
Building intuitions on these objects can take some time, and exercises can take a lot of time to go through at the begining, but don't worry about that. As time goes by things will get clearer :). Also, some of Hartshorne's exercises can be hard for beginners so don't panic if you can't solve all of them.
