Download Basic category theory by van Oosten J. PDF

By van Oosten J.

Show description

Read Online or Download Basic category theory PDF

Best algebra & trigonometry books

Math Word Problems For Dummies

It is a nice ebook for supporting a instructor with constructing challenge fixing quite often. nice rules; stable examples. Mary Jane Sterling is a wonderful author

Fundamentals of Algebraic Modeling: An Introduction to Mathematical Modeling with Algebra and Statistics

Basics OF ALGEBRAIC MODELING 5e offers Algebraic techniques in non-threatening, easy-to-understand language and diverse step by step examples to demonstrate rules. this article goals that can assist you relate math abilities in your day-by-day in addition to various professions together with track, paintings, background, legal justice, engineering, accounting, welding and so on.

Extra resources for Basic category theory

Example text

T X → Y is the composite Now f˜ T (ηX ) T (f ) h T X → T 2X → T Y → Y Since f is a T -algebra map, this is T (ηX ) µX f T (X) → T 2 (X) → T (X) → Y which is f by the monad laws. Conversely, g¯ ˜ : X → Y is the composite ηX T (g) h X → TX → TY → Y By naturality of η and the fact that (Y, h) is a T -algebra, we conclude that g¯ ˜ = g. So we have a natural 1-1 correspondence C(X, U T (Y, h)) T -Alg(F T (X), (Y, h)) and our adjunction. Note that the composite U T F T is the functor T , and that the unit η of the adjunction is the unit of T ; the proof that for the counit ε of F T U T we have that U T εF T T 2 = UT F T UT F T → UT F T = T is the original multiplication µ, is left to you.

Many categories, as we have seen, have a forgetful functor to Set which has a left adjoint. So the forgetful functor preserves limits, and since these can easily be computed in Set, one already knows the “underlying set” of the vertex of the limiting cone one wants to compute. 51 Does a converse to the theorem hold? e. given G : C → D which preserves all limits; does G have a left adjoint? In general no, unless C is sufficiently complete, and a rather technical condition, the “solution set condition” holds.

In this situation, let Q ⊆ P be the image of T , with the ordering inherited from P . We have maps r : P → Q and i : Q → P such that ri is the identity on Q and ir = T : P → P . For x ∈ P , y ∈ Q we have x ≤ i(y) ⇔ r(x) ≤ y (check); so r i and the operation T arises from this adjunction. 1 Algebras for a monad Given a monad (T, η, µ) on a category C, we define the category T -Alg of T algebras as follows: • Objects are pairs (X, h) where X is an object of C and h : T (X) → X is an arrow in C such that T 2 (X) T (h) / T (X) µX  T (X) h  /X h ηX / T (X) Xi ii ii and ii ii h idX "  X commute; f • Morphisms: (X, h) → (Y, k) are morphisms X → Y in C for which T (X) T (f ) / T (Y ) h k  X f  /Y commutes.

Download PDF sample

Rated 4.42 of 5 – based on 28 votes