4 1. Categories and Functors Each arrow has a domain and a target—thus X f Y is a simple diagram with a single arrow f whose domain is X and whose target is Y . If g is another arrow with domain Y , then we can form the ‘composite arrow’ g ◦ f with domain X and target Y . The triangle X f h ◆ ◆◆◆ ◆◆◆ ◆◆◆ ◆◆◆ Y g Z is commutative if h = g ◦ f. If the diagram above is commutative, then we say that h factors through f and through g we also say that h factors through Y . If X and Y are objects in a diagram, we may be able to use the various arrows and their composites to obtain many potentially distinct arrows from X to Y for example, in the cube diagram, there are precisely 6 different composites from X1 to Y4. Each of these paths represents an arrow X1 → Y4, but it can happen that different paths become, on composition, the same arrow. If it turns out that, for each pair X, Y of objects in the diagram, all of the possible composite paths from X to Y are ultimately the same arrow, then we say that the diagram is commutative. We can expand a given (not necessarily commutative) diagram D by drawing as arrows all of the composites of the given arrows, as well as ‘iden- tity arrows’ from each ‘vertex object’ to itself, which compose like identity maps. We’ll refer to the expanded diagram as D. Exercise 1.1. Show that D is commutative if and only if D is commutative. It is frequently helpful to express complicated definitions and properties in terms of diagrams. Here’s an example. Let F be a field, and let F ⊆ E be a field extension. Then there is an inclusion map f : F → E, which just carries an element α ∈ F to the same element, but thought of as being an element of E. Now an algebraic closure for F is an algebraic field extension a : F → A such that for any other algebraic extension f, there is a unique map ¯ making the diagram F a f A E ∃! ¯ commutative. Here you should observe that we use dotted arrows to denote arrows that we do not know exist. Also, this definition gives the algebraic closure as a solution to a ‘universal problem’.

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