2018 - Page 12 of 41 - Shawn Zhong

Category Theory – Video 4

$The Identity Arrow • In S, the Identity Arrow takes each element to itself • e.g. for A={a_0,a_1,a_2 },1_A:A→A≡{█(1_A (a_0 )=a_0@1_A (a_1 )=a_1@1_A (a_2 )=a_2 )┤ • In Music (S), we officially have three more arrows ○ 1_X:X→X ○ 1_Z:Z→Z ○ 1_L:L→L The Identity Laws • Suppose we have a category with objects A and B, then we have 1_A:A→A, 1_B:B→B • The Identity Laws can now be restated to say that, in any Category, this Diagram must Commute • For a diagram to commute, all paths between two objects must be interpreted as the same arrow • Example ○ Suppose f(a_0 )=b_0, then ○ (f∘1_A )(a_0 )=f(1_A (a_0 ))=f(a_0 )=b_0 ○ (1_B∘f)(a_0 )=1_B (f(a_0 ))=1_B (b_0 )=b_0 The Category of Sets With an Endomorphism S^↺ • A^(↺1_A ) and B^(↺1_B ) are objects in S^↺, but what shoud the arrows be? • We want every f:A→B in S to be an arrow f:A^(↺1_A )→B^(↺1_B ) in S^↺ • The law f must follow in S: f∘1_A=1_B∘f, can we generalize this? • What if we replaced 1_A with any endomorphism α, and 1_B with β • Then we get a map f:A^(↺1_A )→B^(↺1_B ) in S^↺ that satisfy: f∘α=β∘f The Associative Law • We have four objects, A,B,C, and D, with arrows f:A→B, g:B→C, and h:C→D • The associativity law says that this diagram must commute • We have 3 paths A→D. They are h∘(g∘f),(hg)∘f, and h∘g∘f • They all must be interpreted as the same map A→D • Example ○ A={a_0,a_1 }, B={b_0,b_1,b_2 }, C={c_0,c_1 },D={d_0,d_1,d_2 } ○ f:A→B≡{█(f(a_0 )=b_0@f(a_1 )=b_1 )┤ ○ g:B→C≡{█(g(b_0 )=g(b_2 )=c_1@g(b_1 )=c_0 )┤ ○ h:C→D≡h(c_0 )=h(c_1 )=d_1$

Category Theory – Video 5

$Calculating The Number of Maps Between Sets • Consider A={a_0,a_1 }, B={b_0,b_1,b_2 } • There are 3×3=9 maps from A to B • There are 2×2×2=8 maps from B to A General Formula for The Number of Maps A→B • In general, the number of maps A→B is |B|^|A| because each a_i has |B| independent choices • The number of maps A→B≡|B^A |=|B|^|A| • In Video 2, we have defined ○ X={x_0,x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_10,x_11 } ○ C={C♮,(C♯/D♭),D♮,(D♯\/E♭),E♮,F♮,(F♯/G♭),G♮,(G♯/A♭),A♮,(A♯\/B♭),B♮} • So |X^C |=|〖12〗^12 |=8,916,100,448,256 Universal Mapping Properties (UMPs) • A Universal Mapping Property asserts something about one or more objects with zero or more maps between them in relation to all other objects X in the Category • A UMP is an object P, that may or may not have a relationship to a diagram of some sort • The principal is that there is only one map between P and all other objects X in that Category • With X as either the Domain or Codomain of that map, such that P still obeys its restriction Initial Object • In any Category C, an object 0 is said to be an Initial Object of C, if ∀X in C, there is a unique C-arrow 0→X • In S, a set 0 is said to be an Initial Object of S, if for all sets X, there is a unique map 0→X • Number of maps A→B is represented by |B^A |=|B|^|A| , so |X|^|0| =1 must hold for all X • So the Initial Set is the Empty Set, because any number raised to the power of 0 is 1 • There are no maps X→0, unless X≅0, since 0 raised to the power of any number (except 0) is 0 Duality • Whenever you define a UMP in S, to get another UMP just "do the opposite of what the definition says" • Usually just by reversing the arrows, and you ll get another UMP for free • Just add the prefix "co-" to the original definition and you’re done Terminal Object • In any Category C, an object T is said to be a Terminal Object of C, if ∀X in C, there is a unique C-arrow X→T • In S, a set T is said to be a Terminal Object of S, if for all sets X, there is a unique map X→T • Number of maps A→B is represented by |B^A |=|B|^|A| , so |T|^|X| =1 must hold for all X • The Terminal Set is 1, because 1 raised to the power of any number is 1 The Opposite Category and Duality • Duality ○ What if for every map f:A→B, we define a map f^op:B→A that makes the same associations between A and B, just from B^′ s perspective? ○ The map f^op:B→A assigns to each b_i∈B a subset A_i of A called the Splitting of b_i, defined as {a∈A_i⇔f^op (b_i )=a}, such that for all a∈A, there is exactly one A_i such that a∈A_i • Opposite Category ○ We can form a new category S^op that is the "mirror", or Opposite Category of S ○ For every object in S, there is a correspondent object in S^op ○ For every arrow f:A→B in S, there is a correspondent arrow f^op:B→A in S^op Calculating the Number of Maps 3→2 in S^op • The number of maps 3→2 in S^op must be 9, because there are 9 maps 2→3 in S • Ideally each element of the Domain in S^op has 3 choices in a map into the Codomain 2 ○ 0: Do nothing ○ 1: Connect ○ 2: Split into 2 • If we want to partition 2 into 3 parts, our options are 2+0+0, and 1+1+0 • We will represent the size of the subset that f^op assigns to b_0,b_1,b_2 as ⟨|A_0 |,|A_1 |,|A_2 |⟩ The Terminal Object in S^op • The Terminal Object in S^op is the Initial Object in S • The Initial Object in S^op is the Terminal Object in S$