Algebraic Geometry Series 1: An introduction to Category Theory

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 \mathcal{C} 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 \iota:\{1\}\to\{1\} is also a legal morphism. The set of morphisms between objects A and B is referred to as Mor(A,B). Morphisms compose as expected: if f\in Mor(A,B) and g\in Mor(B,C), then g\circ f\in Mor(A,C). Composition is associative: f\circ (g\circ h)=(f\circ g)\circ h. Also, for each object A\in Obj(\mathcal{C}), there exists an identity morphism id_A:A\to A. If f\in Mor(A,B), then f\circ id_A=f. Also, if g\in Mor(B,A), then id_A\circ g=g.

Fibered products: These can be best studied with a diagram. Suppose we have \alpha: X\to Z and \beta:Y\to Z, then the fibered product is an object X\times_Z Y such that it is universal for any object A mapping to both X and Y. For example, let us consider the category comprising of all ideals of \Bbb{Z}, and a morphism f:A\to B exists if A\subseteq B. Then, if X=(4) and Y=(6), we can see both lie inside (2). Hence, we can choose Z=(2). Now every object that maps to both (4) and (6) needs to be generated by a multiple of lcm(4,6). Hence, X\times_Z Y will be generated by lcm(4,6), or 12.

Functors: A covariant functor F is like a homomorphism between categories A and B. A contravariant functor G is slightly different. It maps f\in Mor(A,B) to G(f)\in Mor(G(B),G(A)). For example, the identity functor is covariant, while the functor mapping topological open sets (\exists f:A\to B implies A\subseteq B) 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 \Bbb{R}, 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 (0,1) and (0,2). Now consider a differentiable function on (0,2), like x^2. Clearly, it is also differentiable on every subset of (0,2), including (0,1). Hence, there is a mapping from the set of differentiable functions on (0,2) to those on (0,1) (functions map to themselves). However, this mapping may not be surjective. For example, \frac{1}{x-1} is differentiable on (0,1) but not on (0,2).

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer-

4 thoughts on “Algebraic Geometry Series 1: An introduction to Category Theory

  1. 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.

    1. Thanks for the comment Ishan. There might be an infinite number of smooth extensions possible for \frac{1}{x-1.5}. However, I am only talking about \frac{1}{x-1.5}; this exact function, and not another smooth extension of its image in (0,1) This function is clearly not continuous (and hence not differentiabe) over (0,2)

      1. 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.

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