This post is about representable functors:
Definition. Let be a functor. Then is representable if it is isomorphic to for some . In this case, we say that represents .
Exercise. If such exists, then it is unique up to unique isomorphism.
Really one should encode the isomorphism as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation is uniquely determined by the element of corresponding to the identity of .
When is a natural isomorphism, the corresponding element is called the universal object of . It has the property that for every and any , there exists a unique morphism such that .
Example. The forgetful functor is represented by . Indeed, the natural map
is an isomorphism. The universal element is .
Example. Similarly, the forgetful functor is represented by . The universal element is .
A fun exercise (for the rest of your life!) is to see whether functors you encounter in your work are representable. See for example this post about some more geometric examples.
The main example for today is the following:
Lemma. The functor that associates to a topological space its topology is representable.
Proof. Consider the topological space with topology . Then there is a natural map
Conversely, given an open set , we can associate the characteristic function . This gives an inverse of the map above.
The space we constructed is called the Sierpiński space. The universal open set is .
Remark. The space represents the data of open sets for : for any continuous map , we have , where . If denotes the complementary open, then the form a cover of if and only if . This corresponds to the statement that lands in .
Thus, the open cover is the universal open cover, i.e. for every open covering there exists a unique continuous map such that .
Great post, thanks for sharing. I think in your definition of a universal object, it should be:
… It has the property that for every and any , there exists a unique morphism such that .
Thanks, you’re absolutely right! Fixed.