Assume any two vertices are bounded above. There are paths from each that lead to a common vertex higher up.
The colimit of this diagram, if it exists, is called the direct limit, or the injective limit.
Consider the category of sets, partially ordered by inclusion. Morphisms are functions from one set into another, and inclusion is a special kind of morphism.
Consider a set S and create a vertex for each finite subset. Let embeddings act as morphisms among these subsets, and the diagram represents a partial ordering, and it commutes, and every two vertices are bounded above. This system meets our criteria, but smaller systems meet our criteria as well. We don't actually need all the finite subsets of S. The collection must span S, and every pair of vertices must be bounded above. That's all we need. We will show that the set S, represented by the vertex S, along with the embeddings from the finite subsets of S into S, is the direct limit. In other words, S is the colimit of the diagram.
Augment our diagram with another set T. In other words, various functions map our finite subsets into T, and the diagram commutes. We need a function f from S to T that makes everything commute.
Let x be an element of S and start with any finite set containing x, and follow the morphism up to T. Let x map to y in T. Let f(x) = y. But is this well defined?
Suppose there are different finite sets, and in one, x maps to y in T, and in another, x maps to z. Our two finite sets are bounded above by a third finite set, which is part of the diagram. These maps are embeddings, so the map from the third set into T dictates the image of x. It cannot have two different values, hence f(x) = y, and f is well defined.
Does the composite diagram commute? We can follow a chain from x to y in T, or we can follow a chain up to S, which is always an embedding, and maps x to x, then invoke f to get f(x) = y. Yes, the diagram commutes. There is a unique morphism for every T, and S is the colimit, or the direct limit.
A similar proof shows that a module is the direct limit of its finitely generated submodules.
Let's see what goes wrong when pairs are not bounded above. Let M be the abelian group Z3, which is a Z module. Select three finitely generated submodules, each using two of the three generators. Let T = Z, and map our three submodules to T as follows.
[1,0,0] → 1
&
[0,1,0] → 1
[0,1,0] → 2
&
[0,0,1] → 2
[1,0,0] → 3
&
[0,0,1] → 3
We cannot build a group homomorphism from M into T that is consistent with all three maps, hence M is not the direct limit.
To explore the inverse limit, let's move to the category of topological spaces and continuous functions. Remember that arcs, or functions, take you down the lattice. Assume all the spaces in the diagram are disjoint from each other.
Let P be the direct product of the spaces in the diagram. We will build a subspace Q, which inherits its topology from P. Select a morphism in our diagram, which represents a function from U to V. If this function maps a to b, Restrict Q so that the point selected from V = b whenever the point selected from U = a. We have tied two components together. The value of the component U determines the value of the component V. Do this for all functions in the diagram and call the resulting space Q.
We need morphisms from Q down to all these spaces. In other words, Q sits above the diagram and projects downward. In fact, projection is the key. Map Q into V by extracting that particular component, just as we would do with P. If W is an open set in V, its preimage is open in P. This restricts V to W and leaves the other components unconstrained. Restrict this to Q and find an open set in Q, under the subspace topology. This is the preimage of W under the projection from Q to V. Thus the projections from Q down to the various subspaces are all continuous, and are valid morphisms.
Let R be another space with continuous maps to the spaces in our diagram. By assumption, the augmented diagram is still commutative. We need a function f from R to Q that makes everything commute.
Given a point x in R, follow the function from R to V, and let x map to y. This establishes the image of x in P, at least for one of the components. Do this for all components and x has an image in P.
Use the fact that the diagram, with R adjoined, is commutative, to show that f(x) lies in Q.
Is f(R) continuous into Q? If it's continuous into P then it's continuous into Q. A base open set in P restricts finitely many components to open sets within those components. Let x ∈ R be in the preimage. Let V be one of the restricted components; say we are restricting to an open set W in V. Follow the morphism from R directly to V, and x is in the preimage of W. Thus x is in an open set in R. This happens across finitely many components, so take the finite intersection of these open sets in R to produce an open set. Therefore f is continuous, and Q is the inverse limit.
We did not use the fact that pairs of vertices are bounded below; it was not necessary for this example.
As a corollary, give each space the discrete topology. Now every function is continuous, so we are really talking about arbitrary functions among disjoint sets. The inverse limit Q exists, and is a subset of the cross product of the underlying sets, as described above.