On this page, we work work through several of the key examples of limits in the category Set. This is part of a bigger project: Understanding Constructions in Categories.
Recall that a limit, or universal cone, over a diagram is a cone over such that, given any cone , there is a unique cone function? from to .
A terminal object is a universal cone over the empty diagram. In this section, we demonstrate how this leads us to the statement:
Any singleton (a one-element set) is a terminal object in .
To demonstrate, first note that a cone over an empty diagram is just a set and a corresponding cone function is just a function. Therefore, we are looking for a “universal set” such that for an other set , there is a unique function
Singletons fill the bill because for any element we have the unique function defined by
Therefore any singleton is a terminal object in .
A product is a universal cone over a discrete diagram. In this section, we demonstrate how this leads us to the statement:
Any cartesian product is a product in .
To demonstrate, first note that a discrete diagram produces a family of sets with no functions between them. A cone over the discrete diagram consists of a set and a single component
for each set . Therefore, we are looking for a universal cone such that for any other cone , there is a unique cone function
Cartesian product fills the bill because for any , we have the unique function defined by
This function is unique because with any other function
with , then for any element
Therefore .
An equalizer is the universal cone over a parallel diagram
In this section, we demonstrate how this leads us to the statement:
The equalizer of two functions is the subset on which both functions coincide.
To demonstrate, first note that a parallel diagram produces two sets with two parallel functions . A cone over the parallel diagram consists of a set and two components
This is the first example we encounter where the diagram contains morphisms so recall that with a cone we also want the the component diagrams to commute, i.e. we want
Therefore, we are looking for a universal cone such that for any other cone , there is a unique function
Let’s first define the set
with two functions and defined by
and verify that it is indeed a cone. The only thing we need to check is that
which amounts to showing that for all , but that is the definition of , so we do have a cone.
Now let be two cone functions. For any element , we have
so that and the cone function is unique.
Since every cone function is unique, it follows that together with and is a universal cone, i.e. is an equalizer.
A pullback is a universal cone over a cospan
In this section, we demonstrate how this leads us to the statement:
The pullback of two functions of sets is the set of pairs such that .
To demonstrate, first note that a cospan produces three sets with two functions and . A cone over the cospan consists of a set and three components
satisfying
which implies, of course,
Therefore, we are looking for a universal cone such that for any other cone , there is a unique function
Let’s first define the set
with three functions , , and defined by
and verify that it is indeed a cone. The only thing we need to check is that
which amounts to showing that for all , but that is the definition of , so we do have a cone.
Now let be two cone functions with
for functions
For any element , we have
Similarly
so that and the cone function is unique.
Since every cone function is unique, it follows that together with , , and is a universal cone, i.e. is a pullback.
Under Construction fibred products
Last revised on October 9, 2010 at 14:55:06. See the history of this page for a list of all contributions to it.