PROBLEM SET Problem 1.

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1 PROLEM SET 1 PROFESSOR PETER JOHNSTONE 1. Problm Th catgory Mat L. OK, I m not amiliar with th trminology o partially orr sts, so lt s go ovr that irst. Dinition 1.1. partial orr is a binary rlation ovr a st S which is rlxiv, antisymmtric, an transitiv. Dinition 1.2. For two lmnts a, b S o a partially orr st, th mt a b an th join a b o a an b ar rspctivly th gratst lowr an last uppr boun o a an b, whnvr ths things xist. Th way to rmmbr this notation is that th mt is th prouct, an th join is th coprouct, whn arrows go rom small to big lmnts. With that, w can proc to say that th irst part o th problm is wrong. Th suppos morphisms o Mat L o compos associativly (which I m too lazy to vriy, but it s a simpl calculation), but th ntity morphisms rquir th xistnc o lmnts m, M L such that m a M a L. For xampl, th ntity morphism 1 rquirs an lmnt M such that M a = a or all a, i.. a M or all a. I w o a that assumption, w r in, an th ntity morphism at n is an n n matrix with M on th iagonal an m vrywhr ls. Th a is that m srvs as th ntity or aition, an M as th ntity or multiplication Equivalnc to Rl. Hr th analogy is with boolan logic. First, som obsrvations about Rl : th ntity morphisms ar prcisly th minimal rlxiv rlations. W n to xhibit unctors F : Mat L Rl an G : Rl Mat L so that F G an GF ar naturally isomorphic to th ntity. Din F by n {1, 2,..., n} an { (i, j) ij = 1 }. Rcall that our law o composition or rlations is R S = { (a, c) b such that (a, b) R, (b, c) S } which in trms o boolan logic corrspons xactly to th ormula or matrix multiplication; so F is a unctor. In th othr irction, choos an orring o vry st in Rl, an in G by an, th matrix which is a tabl or accoring to th orrs chosn or om an co. gain, by th abov obsrvation, this is a unctor. y our orrly construction o F, w immiatly hav GF = MatL. On th othr han, F G ns som chang o coorinats to gt it to th ntity. This is th momnt to rmark that our choic o orring or vry S ob Rl rally amounts to quipping ach such S with a bijction S : S {1, 2,..., S } = F GS in catgory languag. Now it s clar that th maps { S S ob Rl } giv a natural transormation btwn F G an Rl. 1

2 2 PROFESSOR PETER JOHNSTONE 2. Problm 2. concrt xampl to think o whn aling with mpotnts is projction to a subspac,.g. orthogonal projction in vctor spacs, or som sort o rtraction in topological spacs. concis way to nco an mpotnt is to say that th iagram commuts. concis way to nco a split mpotnt is th commutativ iagram g (i) To think o that thing, w can imagin w v pick a bunch o projctions, an our nw catgory has objcts thr projctions an morphisms maps that prsrv th projctions, i.. maps that ar ssntially in by thir valus on th subspacs onto which w r projcting. Ik what I m rambling about. nyway, lt s vriy th axioms. morphism is a commutativ iagram In, th non-trivial part o that statmnt is =, which ollows by = = = an similarly =. Thn i w hav commutativ iagrams g h C w can stick thm togthr into a nw commutativ iagram, rom which it ollows that hg = hg. Now, or th ntity morphism o :, intuitivly won t work bcaus it has nothing to o with (it osn t know about it). n in you can chck it osn t; what os is, naturally,, an by gnral nonsns, i ntitis xist, thy ar uniqu. (ii) W in I in th obvious way by, ( ) This is clarly a unctor, sinc = or all. It is obviously aithul, an morovr givn any morphism I I in C[Ě], it coms rom th corrsponing morphism in C.

3 PROLEM SET 1 3 What os it man or a unctor T : C D to actor through I? It mans thr xists a unctor T : C[Ě] D such that T = T I. In particular, w hav a commutativ iagram in C[Ě] which gos to a commutativ iagram unr T : T T T g which tlls us that g = T, an morovr on asily chcks g is th ntity bcaus it s th imag o an ntity morphism. From this on asily ss that i T actors thn vry T is split; an vic vrsa, i vry T is split, w can in th actorization T in th obvious way. Th only thing that s not immiat is whr th morphisms o I go unr T ; to gt this, obsrv that vry morphism in I coms rom som morphism in C. W can in th imag o by pulling back to T T T an gluing togthr th maps rom th splitting o an, an this works (haaa i m lazy). (iii) Obsrv that i : (, ) (, ) is an mpotnt in C[Ě], thn is an mpotnt in C. Thn w hav th obvious splitting bcaus = =, = = rom th act that =. Hnc C[Ě] is Cauchycomplt.

4 4 PROFESSOR PETER JOHNSTONE (iv) Suppos (, 1 om ) has an qualizr, as in E Thn w can ool aroun with it..g. obsrv that thr s a uniqu h such that th ollowing commuts: h E Now clarly h =, an morovr, sinc h = =, th map h its as th unlabl arrow in E E but so os E an th map is uniqu, thus h = E. Convrsly, i splits via : X, g : X, it s asy to s that g : X is an qualizr, an th map to X givn a h : is h; vrything works, blalala. Uniqunss ollows asily by th usual shnanigans with arrows. W can play aroun mor an obsrv that w gt a coqualizr rom E: just tak h : E. In, w hav h = hh = h, an by analogous shnanigans w gt uniqunss, tc. Th othr way works similarly. 3. Problm 3. (i) W n to show that or any innr automorphism I an automorphism F, F IF 1 is also innr. Suppos I is isomorphic to via isomorphisms α : I. W claim that th maps F α F 1 : F IF 1 giv that isomorphism. In, ths maps ar isomorphisms, bcaus unctors prsrv isomorphisms, an morovr w hav th commutativ squar F α F 1 F α F 1 F IF 1 F IF 1 F IF 1 bcaus F IF 1 F α F 1 = F (IF 1 α F 1 ) = F (α F 1 F 1 ) = F α F 1 F F 1 = F α F 1.

5 PROLEM SET 1 5 Howvr, thr s a bttr way to s that. Just writ th naturality squar or th morphism F 1 : F 1 F 1 : F 1 F 1 F 1 (ii) α F 1 IF 1 IF 1 IF 1 α F 1 an tak its imag unr F. ut thr shoul b a way to mak it vn mor obvious Exampls class 4.1. Problm 1. la i a la i a la la laa la laaa la la la la. lalala. laaaaa La la la la laa. lalalalaaaaa lala laalala Rgar xistnc o maximal an minimal lmnts in a lattic as part o th inition Problm 2. (i) Notic th quation = is quivalnt to two quations, = =. This shows composition is wll-in. Th ntity o is. (ii) I all ntitis ar in E, in I by I = 1, I( ) = : 1 1. Clarly ull an aithul. For : in E, I() splits as 1 1. ut any actor prsrvs split mpotnts, so i T = T I thn T maps mmbrs o E to split mpotnts. Convrsly, suppos T () splits as T g T T, or all E. Wlog assum th splitting o T ( ) is T 1 T T 1 T T. Thn w v in T on objcts, so that T I = T or all. Givn h : in C[Ě], in T () as th composition g T h. Thn composition works, an in act, vrything works splnly. (iii) lalala (iv) Impotnt splitting is a sl-ual notion, so it s nough to vriy only or qualizrs or coqualizrs. uy on, gt on r!!! lalala I D has qualizrs, consr th unctor ( T T I) : [Ĉ, D] [C, D]. It s ssntially surjctiv on objcts; n to show it s ull an aithul. Th valus o a natural transormation on an objct uniquly trmin th valus o it on any rtract o (an mpotnts ar rtracts), sinc F α G F g F α E F E GE Gg 4.3. Problm 3. (i) lalala. I H is any automorphism, Hα H 1 ntity. is a natural transormation rom HF H 1 to th

6 6 PROFESSOR PETER JOHNSTONE (ii) Evry automorphism F is part o an quivalnc o catgoris, so it will b ull an aithul, so i 1 is trminal, thr s a uniqu morphism F F 1 or any. Sinc it s also surjctiv on objcts, lalala. (iii) St(1, ) is isomorphic to 1 St. F 1 is a singlton, so thr s a uniqu α : 1 St = St(1, ) F. For any st an any a, can rgar a as a map a : 1 ; thn 1 α 1 F 1 a F a commuts. ut F is ull an aithul, so α is bijctiv, so isomorphism, so α is a natural isomorphism. (iv) 1 rprsnts th orgtul unctor Top U St. an 1 is a singlton, so thr s a uniqu natural isomorphism α : U UF. Not that or any X, 1 X an all constant maps ar nomorphisms in Top, so i X has 3 points, it has 4 nomorphisms. blalala, on chcks irctly or all spacs on 2 points. Hnc F (S) = S. What os α S o? Eithr it s a homomorphism or it s not, i.. th map that intrchangs th opn an clos points. ut, or any X, continuous maps X S ar charactristic unctions o opn sts, an by naturality again, X F S α α X F X F F S commuts or any such. Eithr, α X is a homomorphism, or it sns opn substs o X bijctivly to clos substs o F X. Prov thr s an X whos clos sts on t orm a topology in our catgory, th scon cas is rul out, so α must b a natural isomorphism 1 Top F. In bgp, Z rprsnts U : bgp St α S

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