Category Theory Course [Lecture notes] by John Baez

By John Baez

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Need to check: • If [i ] ⊆ [ j] and [ j] ⊆ [k], then [i ] ⊆ [k ] i A X gives f g◦ f j B C g i A X k C • [i ] ⊆ [i ] - easy • If [i ] ⊆ [ j] and [ j] ⊆ [i ], then [i ] = [ j] A i X A i X g f j k B B To show [i ] = [ j], it suffices to show: B A i f B j j g X A g i f i X j B A commute, so i ◦ g ◦ f = i ◦ 1 A and j ◦ f ◦ g = j ◦ 1B , and since i and k are monic, they’re left cancellable: g ◦ f = 1 A and f ◦ g = 1B Next time we’ll define U for subobjects, and this makes Sub( X ), which is a poset (hence a category), into a category with coproducts: ∪ is the coproduct in Sub( X ).

In fact, elt(X) is in 1 − 1 correspondence with the underlying set of X: Given x ∈ X, f : {∗} → X where ∗ → x, and conversely any such f (∗) ∈ X. 4. e. the terminal object in Grp}. So elt(G) has just one element: there’s just one homomorphism f : 1 → G, since 1 is also initial! 5. If C = Cat, elt(D) = {functors f : 1 → D, where 1 is the terminal category in Cat}. functors f : 1 → D are in 1 − 1 correspondence with the objects of D. 1. Suppose C is a category with terminal object 1 ∈ C. Then there’s a functor elt : C → Set with elt( X ) = Hom(1, X ), ∀ X ∈ C and given any morphism g : X → YinC, elt( g) : elt( X ) → elt(Y ) is defined as follows: 1 f X g elt( g) f = g◦ f Y 43 g Proof: elt preserve composition: given X Y h Z we need elt(h ◦ g) = elt(h) ◦ elt( g) f 1 X g Y h Z Given f ∈ elt( X ) we have elt(h ◦ g) f = = = = (h ◦ g) ◦ f h ◦ (g ◦ f ) h ◦ (elt( g) f ) elt(h)(elt( g)( f )) Similarly elt(1x ) f = 1x ◦ f = f , for all f ∈ elt( X ).

G. coproducts, so F (S + T) ∼ = F (S) + F ( T ). Here, S + T is the disjoint union of S and T, F (S + T ) is the free group with elements of S + T as generators, and F (S) + F ( T ) = F (S) ∗ F ( T ) is the “free product” of F (S) and F ( T ) . 2. g. products, so U ( G × H ) ∼ = U ( G ) × U ( H ) where G × H is the usual product of groups G × H. 4. The composite of left adjoints is a left adjoint. The composite of right adjoints is a right adjoint. Proof. Suppose we have functors C adjoint of functors U and U C F D D U U F E and F and F are left E .

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