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This seminar by _llll_ took place on 13th July 2008 20:00 UTC, in #mathematics channel of irc.freenode.net.

The timestamps displayed in the IRC text are UTC+1.
Any explanation or errors regarding this seminar should be recorded in the discussion page.


Topic Edit

Functors and natural transformations, Yoneda's Lemma, adjoints and structures arising from them were among the topics covered.

Seminar Edit

20:58:44 Jafet: _llll_, does this build on last week's?
20:58:57 _llll_: yeah, pretty much

21:00:06 ChanServ changed the topic of #mathematics to: Seminar in progress.
         If you want to ask a question, say ! and wait to be called
21:00:13 _llll_: OK, let's get going
21:00:27 kommodore: hurray!
21:00:29 _llll_: today's topic is functors, natural trnasofmrations and the yoneda lemma.
21:00:39 _llll_: monads may be mentioned if there is time at the end
21:01:03 _llll_: OK, so last time i defined category and said that pretty much everything
                 forms a category
21:01:19 _llll_: so it's reasonable to assume that "categories" themselves form a category
21:01:35 _llll_: ie, given categories CC and DD there should be a notion of map F:CC-->DD
21:01:43 _llll_: such an F is called a "functor"
21:02:03 _llll_: the definition is what you'd expect,  F sends an object A in CC to an object
                 FA in DD
21:02:18 _llll_: and if f:A-->B is a map in CC, then it gets mapped to Ff:FA-->FB in DD
21:02:35 _llll_: and if g:B-->C is another map in CC then Ff # Fg = F(f#g)
21:02:58 _llll_: (so recall that im writing composition of f:A-->B followd by g:B-->C as f#g :A-->C
21:03:13 _llll_: the final condition is that F(id_A)=id_FA ie identities get apped to identities
21:03:55 _llll_: composing functors is just like composing functions, and there is an identity
                 functor id_CC : CC-->CC which just sends an object to itself (and similarly
                 on morphisms)
21:04:09 _llll_: so we get a category CAT where the objects are categories and the maps are functors
21:04:16 ness: !
21:04:20 _llll_: go ahead ness
21:04:35 ness: are there functors between every two objects in CAT?
21:05:01 _llll_: not necessarily, CC can have no objects and no arrows (the empty category)
21:05:11 _llll_: so then there are no functors DD--> "the empty category"
21:05:21 ness: oh
21:05:37 ness: thanks
21:05:39 _llll_: but otherwise you can just pick X in DD and send every object to X and every map to the identity on X
21:06:13 _llll_: (oh, well, if DD is also "the empty category" then there is actually a map from
                  "the empty category" to itself, namely the identity functor)
21:06:16 _llll_: but, anway
21:06:51 _llll_: examples of functors are sending a group G to the set of elements of G
21:07:16 _llll_: this is U:Grp --> Set the "forgetful functor" -- you forget that G is a group,
                 ie UG is just a set
21:07:43 _llll_: going the other way, from a set X you can form the free group on X, this gives
                 a functor F:Grp --> Set
21:08:06 _llll_: so now suppose that we have two functors F and G which both go CC-->DD
21:08:21 _llll_: we can define a map F-->G
21:08:30 _llll_: ie something that goes between *functors*
21:08:54 _llll_: this is called a "natural transformation"
21:09:14 _llll_: the definition is that m:F-->G means that for every X in CC we get a map
                 m_X FX-->GX in DD
21:09:49 _llll_: such that whenever f:A-->B is a map in CC, m_A # Gf = Ff # m_B
21:10:03 _llll_: this equation is best pictured as a square, which i will now attempt to draw

,----[ a picture of what it means for m:F-->G (pasted to pastebin.ca during the seminar ]
|       Naturality condition
|       ====================
|        
|       Let F and G be functors CC-->DD.
|       Then "m:F-->G is a natural transform" means that for all A in CC
|        
|       m(A) : FA-->GA in DD
|        
|       and for all  f:A-->B in CC,
|        
|          FA --m(A)--> GA
|           |           |
|         Ff|           |Gf     (1)  "Naturality square for m at f"
|           |           |
|           v           v
|          FB --m(B)--> GB
|        
|       commutes, i.e.,
|        
|         m(A) # Gf = Ff # m(B)   (2)
|        
|       for all f:A-->B in CC
|        
|       Equvialently, let [1] be the poset {0<=1} considered as a category.
|       Then m is "the same" as a functor
|        m:CCx[1] -->DD is
|       with
|         CC--i_0-->CCx[1]--m -->DD = F
|         CC--i_1-->CCx[1]--m -->DD = G
|        
|       where i_0(A)=(A,0) and i_1(A)=(A,1) -- ie a natural transformation is
|       a homotopy.
|        
|       Remark 1:  There are functors Ner and |-|
|        Cat--(Ner)-->SSet--(|-|)-->Top
|        
|       which turn natural transformations into real homotopies.  Here SSet is
|       the category of simplicial sets.
|        
|       Remark 2: The map equal to either side of equation (2) is the diagonal
|       of the square (1).  We can write it as m^f : FA-->GB then we have
|       split the square into two triangles -- this is how homotopies of
|       simplicial sets work.  ( m^f=m(f,0<=1) with the previous formulation)
`----

21:11:52 _llll_: so you can compose natural transforms in (hopefully) the obvious way
21:12:17 ~lxuser: !
21:12:20 _llll_: so there is a category [CC,DD] where the objects are functors CC-->DD
                 and the maps are natural transformations
21:12:23 _llll_: go ahead lxuser 
21:13:06 ~lxuser: can you form functors between a category and itself?
21:13:36 _llll_: sure, F:CC-->CC is always possible, there is one fucntor called
                 "the identity on CC" which sends A in CC to A
21:13:40 _llll_: and f:A-->B to f
21:14:18 _llll_: (a monad is a special kind of functor from a category to itself)
21:14:41 _llll_: so to illustrate natural transformations a bit more, im going to prove
                 (a version of) "the yoneda lemma"
21:14:48 _llll_: this is about functors CC-->Set
21:15:15 _llll_: ie we fix a category CC and consider the category [CC,Set] of functors
                 from CC to Set
21:15:37 _llll_: one example is the functor CC(A,-)
21:15:54 _llll_: this notation is meant to suggest that A is fixed in CC, and that - is a
                 "placeholder"
21:16:00 _llll_: CC(A,-):CC-->Set
21:16:12 _llll_: it sends objects B in CC to the set CC(A,B)
21:16:21 _llll_: ie to the set of maps from A to B in CC.
21:16:43 _llll_: if f:B-->B' then CC(A,-) has to send f to a function from CC(A,B) to CC(A,B')
21:16:50 _llll_: which function should it be?
21:17:04 _llll_: the obvious way is to send g:A-->B to g#f
21:17:19 _llll_: (since g#f goes A-->B' this ends in the right place)
21:17:32 _llll_: it's easy to check that C(A,-) really is a functor
21:18:34 _llll_: The yoneda lemma says that if F:CC-->Set is any other functor, then there is
                 a natural bijection between natural tramsformations CC(A,-) to F and
                 the set FA
21:18:55 _llll_: it's actually really easy to prove, once you get the notation straight
21:19:02 _llll_: so what does F:CC-->Set look like?
21:19:15 _llll_: so if A is an object in CC, FA is some set
21:19:28 _llll_: and if f:A-->B then Ff is a function FA-->FB
21:19:37 thermoplyae: !
21:19:42 _llll_: go ahead thermoplyae 
21:19:47 thermoplyae: does the word 'natural' in 'natural bijection' have meaning?
21:19:54 _llll_: ah, yes
21:20:18 _llll_: it means that actually it is part of a natural transformation which is an
                 isomorphism
21:20:32 _llll_: the bijection if [CC,Set](CC(A,-),F) ~ FA
21:20:39 _llll_: and it is natural in both A and F
21:20:54 _llll_: so *actualyl* there are functors  CC^op x [CC,Set] --> Set
21:21:02 _llll_: one sends (A,F) to [CC,Set](CC(A,-),F)
21:21:08 _llll_: and the other sends (A,F) to FA
21:21:27 _llll_: and these two are naturally isomorphic in the category
                 [CC^op x [CC,Set],Set] !
21:21:46 _llll_: (a bit complicated and not terribly important to understand at this stage)
21:22:15 _llll_: ok, so as i was saying
21:22:22 _llll_: if f:A-->B then Ff is a function FA-->FB
21:22:42 _llll_: write a^f for the element Ff(a)
21:22:52 _llll_: ie if a in FA then a^f is in FB
21:23:23 _llll_: the statement "F is a functor CC-->Set" means that if f:A-->B and g:B-->B'
                 and a in FA, then 
21:23:32 _llll_: (a^f)^g = a^(f#g)
21:23:40 _llll_: (this is equality in the set F(B'))
21:23:47 _llll_: and also a^{id_A} = a
21:24:15 _llll_: if you know about groups acting on sets (or groups, or similar) then this is
                 exactly the same rules as there
21:25:05 _llll_: ok, so now what does it mean for m to be in [CC,Set](CC(A,-),F)?
                 (ie m:C(A,-) --> F  natural transofmration)
21:25:28 _llll_: it means that for every B in CC, we have a function m_B : CC(A,B) --> FB
21:26:10 _llll_: such that whenever f:B-->B' (map in CC), m_B # Ff = CC(A,-)(f) # m_{B'}

,----[ a picture of m: C(A,-) --> F posted to pastebin.ca during the seminar]
| Theorem (Yoneda Lemma)  [CC,Set](CC(A,-),F) ~ FA
|  (an isomorphism of sets natural in F and A)
| Proof:
|  First some notation.
| 
|  If f:B-->B' then Ff:FB-->FB' is a map of sets, use the notation Ff(x)=x^f
| 
|  So F is a functor iff (x^g)^f = x^{g#f} and x^{id_A}=x
| 
|  [e.g., CC(A,-):CC-->Set is the functor with g^f=g#f]
| 
|  The m is a natural transformation CC(A,-) to F iff the following squares commute
| 
|    g|--------------------------> m_B(g)
|    -                               -
|    |   C(A,B) --- m_B -----> FB    |
|    |     |                   |     |
|    |     | C(A,f)            |Ff   |
|    |     |                   |     |
|    |     v                   v     |
|    |   C(A,B') -- m_{B'} --> FB'   |
|    |                               v
|    |                              m_B(g)^f
|    v                                  
|   g#f |-------------------------> m_{B'}(g#f)
| 
|  I.e., m is natural iff
| 
|    m_B(g)^f=m_{B'}(g#f)  (*)
| 
|  for all g:A-->B and f:B-->B'.
| 
|  Put B=A and g=id_A (f:A-->B' is still an arbitrary map from CC).
|  Then (*) becomes
| 
|    m_A(id_A)^f=m_{B'}(f)
| 
|  i.e., m is determined by the single element m_A(id_A) in FA.
|  And conversely, given a in FA, define m:C(A,-)-->F by
| 
|    m_B(g)=a^g
| 
|  which gives m:C(A,-)-->F natural transformation.
| 
|  So the proof is simply to note that
| 
|    [C(A,-),F] --> FA
|       m |--------> m_A(id_A)
| 
|      a^(-) <------|a
| 
|  are a pair of mutually inverse bijections.  These are natural
|  in F and A.                                                       QED
`----


21:27:31 _llll_: if you unpick what this equation means, starting from g in CC(A,B), the
                 function 'm_B # Ff ' sends this to Ff(m_B(g)) in FB
21:27:40 _llll_: ie to m_B(g)^f
21:27:51 _llll_: using my "action" notation from before
21:28:17 _llll_: and the other side CC(A,-)(f) # m_{B'} sends g to m_{B'}(g#f)
21:28:32 ~carolyn_: !
21:28:44 _llll_: so m is natural means that m_B(g)^f=m_{B'}(g#f) for all g:A-->B and
                 f:B-->B' (f and g are maps in CC)
21:28:47 _llll_: go ahead carolyn_ 
21:29:11 ~carolyn_: what is "commute" in the proof / picture
21:29:21 _llll_: oh, good point
21:29:32 _llll_: it just means that the 2 different ways of going round the diagram are equal
21:29:50 _llll_: so for a square, if you start at the top-left you can either go right then down
21:29:54 _llll_: or down then right
21:30:11 _llll_: both ways give you a composite, and these must be equal
21:30:45 _llll_: so looking at http://pastebin.ca/1071102, the square commutes just means that
                 m(A) # Gf = Ff # m(B)
21:31:12 _llll_: ok, going back to m: CC(A,-) --> F
21:31:24 _llll_: i said that m is natural iff m_B(g)^f=m_{B'}(g#f) for all f,g
21:31:39 _llll_: but now what happens when we put B=A and g=id_A 
21:31:49 _llll_: we get m_A(id_A)^f=m_{B'}(f)
21:31:58 _llll_: but this actually determines all of m
21:32:16 _llll_: sice to define m we just need to define m_{B'} for all B'
21:32:29 _llll_: so m is determined by this one element m_A(id_A) in FA
21:32:36 _llll_: conversely, if a is any element in FA
21:33:11 _llll_: define a n_B: CC(A,B) --> FB by n_B(g) := a^g
21:33:20 _llll_: ie n_B(g) = Fg(a)
21:33:37 _llll_: it's not hard to check that n is then a natural transformation CC(A,-) to F
21:33:52 _llll_: ie the bijection between [CC(A,-),F] and FA
21:34:03 _llll_: sends m to m_A(id_A)
21:34:20 _llll_: and the inverse sends a to this n with n_B(g)=a^g
21:34:50 _llll_: so that pretty much completes the proof, apart from the "natural in A and F"
                 bit, which im going to ignore/leave as an exercise
21:35:03 _llll_: so that was "the yoneda lemma"
21:35:34 _llll_: The next part of the seminar is about adjunctions
21:36:13 _llll_: suppose F:CC-->DD and U:DD-->CC are two functors
21:36:46 _llll_: "F is left adjoint to U" means that there is a bijection between
                 DD(FX,Y) and CC(X,UY)
21:37:05 _llll_: and this has to be "natural in X and Y", which again i wont worry about too much
21:38:02 _llll_: so this means that if f:FX-->Y in DD we get a corresponding f^ : X-->UY in CC
21:38:17 _llll_: and if g:X-->UY in CC then we get a g^: FX-->Y in DD
21:38:26 _llll_: and f^^=f and g^^=g
21:38:58 _llll_: the proof of the yoneda lemma should suggest that everything is determined by
                 what happens when f=id_{FX} and g=id_}{UY}
21:39:40 _llll_: ie id_FX: FX-->FX gets sent to some map unit_{X}:X-->UFX
21:39:58 _llll_: and id_{UY} gets sent to some map counit_Y : FUY --> Y
21:40:43 _llll_: it turns out that these maps actually determine the whole of the adjunction
                  -- that comes from the "natural in X and Y" part, but it's actually just the
                 yoneda lemma proof again
21:40:47 _llll_: namely
21:41:16 _llll_: if f:FX-->Y then f^ is equal to the composite unit_X # F(f)
21:41:33 _llll_: ie X--unit_X--> UFX -- U(f) -->  UY
21:41:53 _llll_: and g^ = (FX-- Fg --> FUY -- counit_Y --> Y)
21:42:28 _llll_: the statement that f^^=f then tells us that f is equal to the composite
21:42:58 _llll_: FX --F(unit_X) --> FUFX -- FU(f) ---> FUY -- counit_Y ---> Y
21:43:13 _llll_: ie f= F(unit_X) # FU(f) # counit_Y
21:43:36 _llll_: it turns out that counit is a natural transformation from U#F to the identity
21:43:50 _llll_: ie FU(f) # counit_Y = counit_{FX} # f
21:44:14 _llll_: so "f^^=f" becomes  F(unit_X) # counit_{FX} # f = f
21:44:23 _llll_: for all f:FX-->Y
21:44:37 _llll_: so when f=id_FX we get F(unit_X) # counit_{FX} = id_{FX}
21:45:13 _llll_: and similarly, from g:X-->UY and g^^=g we get the equation
                 unit_{UY} # U(counit_Y) = id_{UY}
21:45:30 _llll_: these two equations F(unit_X) # counit_{FX} = id_{FX} and
                 unit_{UY} # U(counit_Y) = id_{UY} are called the "triangle identities"
21:45:49 _llll_: if unit_X and counit_Y are any natural transformations satisfying those, then
                 F and U are adjoint
21:45:59 _llll_: ok, so, why do we care?
21:46:24 _llll_: the main reason is that if F is adjoint to U then it is a theorem that F preserves
                 colimits and U preserves limits
21:47:01 _llll_: one example of an adjunction is that F = free group functor Set --> Group and
                 U=forgetful functor , U:Group-->Set
21:47:29 _llll_: so F preserves colimits
21:47:33 _llll_: and U preserves limits
21:47:50 kommodore: !
21:47:57 _llll_: go ahead, kommodore 
21:48:32 kommodore: Do you really mean F is adjoint to U, or F is left adjoint to U?
21:48:47 _llll_: F is left adjoint to U
21:48:51 _llll_: U is right adjoint to F
21:49:13 kommodore: ok.
21:49:14 _llll_: you can remember because in the bijection DD(FX,Y) ~ CC(X,UY)
                 F is on the left and U is on the right
21:49:31 _llll_: so the result is that left adjoints preserve colimits, and right adjoints
                 preserve limits
21:49:48 _llll_: even for groups, this is perhaps not totally obvious
21:50:33 _llll_: if X is a set, X= coproduct_{x in X} {x} so
                 Free(X) = coproduct_{x in X} F{x} says every free group is a coproduct of
                 infinite cyclic groups
21:50:50 _llll_: so to quickly talk about monads
21:51:01 _llll_: suppsoe F is left adjoint to U

[nb: in the smeinar I messed up even the definition of T, luckily
 FunctorSalad put me right.  I have therefore made some corrections
 below -- _llll_ ]

21:51:11 _llll_: then T=F # U is a functor CC-->CC
21:51:21 _llll_: it sends X to TX=UF(X)
21:51:32 _llll_: i said the adjunction is determined by the unit and counit
21:51:39 _llll_: unit_X : X--> UFX
21:51:45 _llll_: so unit : id --> T
21:51:59 _llll_: (and this is actually a natural transformation from the identity to T)
21:52:13 _llll_: the counit goes conunit_Y:FUY-->Y
21:52:31 _llll_: so therefore U(counit_{FX}) : UFUFX-->UFX
21:52:55 _llll_: ie if we put m_X=U(counit_FX} then m:TTX-->TX
21:53:27 _llll_: and the triangle identities say that m satisfies the equation
                  T(m_X) # m_X = m_{TX} # m_X
21:53:44 _llll_: if we think of the F=free-group / U=forgetful functor stuff
21:54:45 _llll_: then X is a group and TX is the free group on the elements of X
21:54:55 _llll_: so this is like putting the elements of X in boxes
21:55:03 _llll_: which i suppose is why functional programmers like using monads
21:55:49 _llll_: formally, a monad on CC is a functor T:CC-->CC with natural transformations
                 unit_X : id-->T and m: T^2 --> T
21:56:14 _llll_: satisfying T(m_X) # m_X = m_{TX} # m_X and
                 unit_{TX} # m_X = id_X and T(unit_X) # m_X = id_X

[_llll_ getting confused elided]

21:56:14 FunctorSalad_: !
21:56:18 _llll_: go ahead FunctorSalad_ 
21:56:28 FunctorSalad_: I think for the monad you need to start with a *set* X
21:56:35 FunctorSalad_: the other way would be a comonad

21:58:29 _llll_: so what is TX, it's the set of formal words in elements of X
21:58:40 _llll_: so it's still putting elements of X into boxes
21:59:08 _llll_: ah, yeah this way makes much more sense
21:59:29 _llll_: because an *algebra* for a monad is a map TX--a-->X
21:59:36 _llll_: TX is formal words in elements of X
21:59:53 _llll_: a says how to identify some of these words
22:00:14 _llll_: every group can be given by generators an drelations, and this is actually
                 the same thing
22:00:22 _llll_: ie you can define a category Alg(T) for a monad
22:00:58 _llll_: when the monad comes from an adjunction, if CC is is equivalent to Alg(T)
                 then we say the adjunction is "monadic"
22:01:24 _llll_: this means that the elements of CC (so groups here) are, in some sense,
                 given by genberators and relations using the adjunction
22:01:45 _llll_: there's a theorem called the Beck monadicit theorem that makes this more formal
22:01:58 _llll_: but i think this is a good place to stop
22:02:05 |Steve|: !
22:02:11 _llll_: time for questions
22:02:14 _llll_: go ahead |Steve| 
22:02:23 |Steve|: Alg(T) is the algebra for a monad? What does that mean?
22:02:32 _llll_: Alg(T) is the cateory of algebras for the monad
22:02:46 _llll_: so the objects are TX--a-->X
22:03:47 _llll_: presumably maps a-->b are f: X-->Y such that
                  TX--a-->X-->Y = TX--T(f)-->TY--b-->Y
22:04:41 _llll_: if you can prove that some category is monadic over Set then there are a load of
                 theorems about limits in that category that help you compute things
22:05:00 |Steve|: Is that first diagram TX--a-->X--f-->Y?
22:05:08 _llll_: yeah
22:05:11 |Steve|: Okay, thanks.
22:07:12 _llll_: OK, I guess that ends the seminar
22:10:50 _llll_: Next week somiaj is going to talk about measure theory

22:15:32 ChanServ changed the topic of #mathematics to: For help go to #math | NEXT SEMINAR: Sunday 20th July
         at 20:00 UTC: Introduction to Measure Theory by somiaj | Later Seminars: see
         http://freenode-math.com/index.php/Seminars

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