### Algebraic Geometry Series 1: An introduction to Category Theory

#### by ayushkhaitan3437

This is intended to be a series of articles, that fleshes out the bare skeleton of Algebraic Geometry. I will be closely following Ravi Vakil’s treatment of the topic, supplemented with other treatments.

**A category consists of a “collection” of objects, and for each pair of objects, a set of morphisms (or arrows) between them.**

Note that the objects being considered need not be distinct. In the category of sets, for instance, the identity morphism is also a legal morphism. The set of morphisms between objects and is referred to as Mor. Morphisms compose as expected: if Mor and Mor, then Mor. Composition is associative: . Also, for each object Obj, there exists an identity morphism id. If Mor, then id. Also, if Mor, then id.

**Fibered products**: These can be best studied with a diagram. Suppose we have and , then the fibered product is an object such that it is universal for any object mapping to both and . For example, let us consider the category comprising of all ideals of , and a morphism exists if . Then, if and , we can see both lie inside . Hence, we can choose . Now every object that maps to both and needs to be generated by a multiple of . Hence, will be generated by , or .

**Functors**: A covariant functor is like a homomorphism between categories and . A contravariant functor is slightly different. It maps Mor to Mor. For example, the identity functor is covariant, while the functor mapping topological open sets ( implies ) to differentiable functions on them is a contravariant functor.

I would like to discuss the contravariant functor mentioned above. Let us consider the category of open sets in , such that morphisms exist between two objects only when the domain is a subset (not necessarily proper) of the range. For example, there exists a morphism between and . Now consider a differentiable function on , like . Clearly, it is also differentiable on every subset of , including . Hence, there is a mapping from the set of differentiable functions on to those on (functions map to themselves). However, this mapping may not be surjective. For example, is differentiable on but not on .

Just a minor comment : The example you have given in the last line should be replaced by something like 1/(x-1) as the function 1/(x-1.5) on (0,1) does find a smooth extension to the interval (0,2) (though the extension is not given by the same equation of course) and hence lies in the image.

Thanks for the comment Ishan. There might be an infinite number of smooth extensions possible for . However, I am only talking about ; this exact function, and not another smooth extension of its image in This function is clearly not continuous (and hence not differentiabe) over

Thanks Ayush, but I am not sure I understand this correctly. Below I write some statements according to my understanding, please let me know if I am making some mistake :

1.) Let i : (0,1) \rightarrow (0,2) be the inclusion map. Let F=Hom(-,R) . We get an induced map F(i) : Hom((0,2),R) \rightarrow Hom((0,1),R) . You wish to show that F(i) is not surjective.

2.) The function f:(0,1) \rightarrow R defined by f(x)= \frac{1}{1.5-x} is well defined on this interval and is differentiable (in fact smooth)

3.) f finds a smooth extension g to the interval (0,2). But g can not be given by the same expression on the interval (0,2). g restricts to f on the interval (0,1). Thus (F(i))(g)=f and hence f lies in F(i)(Hom((0,2),R)). Thus this example can not be used to show that the map F(i) is not surjective.

4.) The function h(x)= \frac{1}{1-x} on (0,1) ,on the other hand, does not find a differentiable extension. Thus can not be obtained by restricting a well defined differentiable function on (0,2). This shows that F(i) is not surjective.

Thanks

Ishan after our convesation on facebook, it seems to me that you’re right. I have corrected the error in the main post. Thank you! As always, your comments are invaluable.