# Category Theory Course [Lecture notes] by John Baez

By John Baez

Best quantum theory books

Introduction to the theory of quantized fields

During this variation we've got rewritten the chapters that debate the tools of continuing integration and the renormalization crew, that are subject matters in conception that experience turn into vitally important in recent times. we've additionally remodeled and supplemented the sections at the entire eco-friendly features.

Quantum inverse scattering method and correlation functions

The quantum inverse scattering approach is a method of discovering special recommendations of two-dimensional versions in quantum box conception and statistical physics (such because the sine-Gordon equation or the quantum nonlinear Schrödinger equation). This creation to this crucial and interesting zone first bargains with the Bethe ansatz and calculation of actual amounts.

A First Course in Group Theory

One of many problems in an introductory e-book is to speak a feeling of function. in basic terms too simply to the newbie does the booklet turn into a series of definitions, suggestions, and effects which look little greater than curiousities best nowhere specifically. during this ebook i've got attempted to beat this challenge by way of making my significant target the selection of all attainable teams of orders 1 to fifteen, including a few learn in their constitution.

Factorization Method in Quantum Mechanics

This booklet introduces the factorization technique in quantum mechanics at a sophisticated point, with the purpose of placing mathematical and actual thoughts and methods just like the factorization procedure, Lie algebras, matrix components and quantum regulate on the reader’s disposal. For this goal, the textual content offers a accomplished description of the factorization technique and its large functions in quantum mechanics which enhances the normal assurance present in quantum mechanics textbooks.

Extra info for Category Theory Course [Lecture notes]

Example text

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 .