# 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 $(0,2)$.

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

1. ishanmata says:

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. ayushkhaitan3437 says:

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. ishanmata says:

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

2. ayushkhaitan3437 says:

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.