Categories in recursion theory,

b y , y by

Michael G. Wasserman B.A., Williams College

(1968)

SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE

DEGREE OF

DOCTOR OF PHILOSOPHY at the

MASSACHUSETTS INSTITUTE OF TECHNOLOGY September 1971

Signature of

Certified by

Accepted by.

Author.

Department o "--iath tis, July, 971

'jThesis upervisor

Chairman, partmental Committee Archives on Graduate Students

SEP 30 1971

4-8R AR\ES

Signature redacted

Signature redacted

Signature redacted

C.

Categories in Recursion Theory by

Michael G. Wasserman

Submitted to the Department of Mathematics in July, 1971 in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Mathematics.

ABSTRACT

This dissertation presents an extensive snapshot of elementary recursion theory taken through the lens of category theory. It is divided into five chapters.

In the first chapter, several basic categories

implicit in recursion theory are made explicit,, and their properties are explored. _{In the second chapter,} recursion theoretic notions are expressed in terms of properties of the basic categories. In the third

chapter, some important recursion theoretic results are proved in strictly categorical language, using several simple axioms about the basic categories.

The fourth chapter extends the results of the first three chapters to the analogous categories implicit in generalized recursion theory. The basic categorical distinction seems to be in ordinary recursion theory, the notions "finite," "la-finite," and "bounded" coincide, while in generalized recursion theory they are distinct.

Finally, in Chapter 5, a categorical formulationcf Turing reducibility is given, and, within this formulation,

a basic theorem about the jump is proved.

Thesis Supervisor: Hartley Rogers, Jr. Title: Professor of Mathematics

Acknowledgements

Mathematical acknowledgements go to my colleagues for their intellectual stimulation, and to Hartley Rogers for suggesting this topic and otherwise helping to get the ball rolling. _{My deepest thanks go to my wife,} however, for standing by me through the darkest days,

4.

Table of Contents

Chapter

I. The Basic Categories and Their Properties

1.

Rl,

R

₂ and

R1

2. Coproducts, Products, and Pullbacks 3. A Question of Balance

4. _{Images, Coimages, Reversible Monomorphisms,} Quasi-Unions, and Essential Domains

5. Singletons 6. Complements 7. _{The Ordering <C}

II. Recursion Theoretic Properties Expressed

Categorically

1. _{Characterizations of}

N

2. Facts About R. E. Objects 3. Recursive Monomorphisms

4. The Notions Simple, Hyperhypersimple, and Maximal

5. Indices; Productive Objects 6. The Halting Problem

III. Categorical Proofs

1. More About Singletons

2. The Pullback Criterion

Page 7

7

9

23 30 38 41 4)4

58

58 66 70

76

83 93 96 96 108

..- , a

Chapter

III. 3. Product Maps

4. Essential Domains and R. E. Objects 5. More About Recursive Monomorphisms 6. Many-one Reducibility

7. Three Theorems

8. _{The Recursion Theorem, Myhill's Theorem,} and Creative Monomorphisms

IV. Extension to Admissible Ordinals 1. The Categories and Simple Axioms

2. The Notion of a-R. E. 3. The Remaining Axioms

4. Projectibility and Boundedness

5. Extended Notions: a-Cohesive, a-Immune V. Reducibility Systems

1. Definition 2. The Functors I

3. A Completeness Theorem 4. S-Reducibility

5. Least Upper Bounds 6. A Theorem on Jumps Appendix Page 110 115 140 156 167 175 189 189 199 207 211 218 228 228 231 236 238 239 245 251

Categories in Recursion Theory

Note: The first four chapters of this dissertation build

up progressively to the definition of "recursive category"

actually made at the beginning of Chapter V. The development

proceeds by mentioning various categorical axioms;

eventually a recursive category is defined to be one

that satisfies all the axioms mentioned.

At the same time, certain very specific categories,

arising out of recursion theory, are under study. The

axioms were designed to capture the properties of these

categories, and likewise, the categories in question had

better satisfy the axioms; occasionally this required proof.

Thus there arises in this dissertation the curious syntax:

Axiom-proof. The proof in such a case is a proof not of the axiom, of course, but of the statement: "The category under study satisfies this axiom."

Finally, many of the results in the first four chapters are labelled with a star. Such results are proven using only the axioms and general facts about categories; that is, they hold pot only of the specific category in question, but in general for all recursive categories.

Chapter V defines and examines the notion of "reducibility

system." The treatment is totally abstract with no specific

7.

Chapter I

The Basic Categories and Their Properties

1.

Rl,

R

₂

,

and

R'.

1.1. Notation. _{We shall work, at the outset, with} the natural numbers N = {0, 1, 2,...} plus a special symbol c. Our basic objects will be subsets of

N I) {co}. _{In general, such subsets will be denoted}

A, B, ... A will denote A - {oo}. On the other hand,

00

we may wish to begin with a subset of N. Such subsets will in general be denoted X, Y, ... ; X will denote

X U {oo}. We shall also use the symbol N to denote

00

N = N ( {a}.

We adopt the following notations for special morphisms:

C-* monomorphism

++ epimorphism +~ isomorphism.

1.2. _{Definition. Category}

_Rl*

_{The objects of}

_R

1

are subsets of

I,

containing G. A morphism _{a : A+ B,} in

Ri

is afunction _satisfying:

8.

()

3

a partial recursive function $ such that

= n) E B if $(n) is convergent

for all n e A,

a(n)

O

if $ is _{divergent at n.}

(i) a(cO) = 00.

We shall say that $ induces a. Given a partial recursive function 4, define 4* : N + N via

$ (a) if convergent S=1if divergent.

We note that

R,

_{is a small cetagory, though this} is not essential to our purposes. _{Note also that a} morphism in

Rl

_{is a monomorphism iff it is 1-1, and}

an epimorphism iff it is onto.

1.3. Axiom.

R,

has a 0 object.

Proof: {w} clearly serves. We note in passing that the 0 of

R,

is unique, and not merely unique up to isomorphism,

1.4. Axiom.

R,

has kernels.

Proof: Let a : A -+ B be given. Let

9.

inclusion. Clearly i serves as ker (a).

1.5. Definition. Categories

R

2

,

Rt.

R

2 is the

subcategory

R,

including all objects, but only morphisms with kernel 0.

The objects of

R

are subsets of

a : X + Y is a function such that

3

recursive function $ with X C domain

N. A morphism a partial

(4) and

$(x) = a(x) e Y for all x e X.

Clearly the categories

R

₂ and R are naturally

isomorphic. Note that

R2

is definable from

Rl,

so

all statements about R and R will be made for R

though proofs may occasionally be carried out for

R'.

1.6. Axiom. _{0 serves as a conull object for}

_R2;

R2

also has a null object.

Proof: Any singleton {n) serves as a null

object for

R',

i.e. {n, o} is null for R₂

.

We shall

refer at times to R₂-null objects {n, co} as singletons.

2. Coproducts, Products, and Pullbacks.

in fact, given A and B, the same pair of morphisms A + C, B -+ _{C serves as the coproduct in}

_R

1 and in R2' Proof: _{Recall that, for X, Y C N,}

X join Y = {2xlx e X}

U

{2x+l|x E Y}. _{Given Al, A2,} objects in

Ri,

let C = (A₁ join A2)w. We have

00 : natural injections u i A AiC-* C via: 2x if x w u(x) f x 2x+l if x

$

w and u₂(W _{=t = 00} Suppose

$ni (i1, 2) induce maps Ai + B. Then the unique compatible map C -+ B is induced by

n (c/2) if c is even 0(c)= nl

$n (c ) if c is odd. s 2 A@

Thus (C, ul, u₂)

serves as A 1 A2 in Ri. Note that ul and u2

have kernel 0, and that if the maps A -+ B have

ker 0, then so does the map C -+ B, so (C, u , u2)

also serves as A₁ @ A₂ in

R

₂

_.

2.2. Lemma. If A++ B, then B is recursively

00

enumerable in A.

IL i *

Proof: Say A -+ B is induced by $n. Define

e(x,

A) via: dovetail computations for n (y)

W0

for all

y s A. If one eventually converges to x, give output; otherwise diverge. Then B =

{x|6(x,

A) convergent}.

00 00

2.3. Remark.

R,

does not admit arbitrary infinite coproducts; similarly for R₂*

Proof: Let Ai = {i, cc} for i e N. Suppose

C = @ Ai, with injections f : Ai C. Let x be any infinite subset of N, say X = {x1 , x22 *...}

Let g :Ai +X via f 1 Then 3!h : C + X

such that for all i, hfi = g In particular,

C0

h : C++ X, so X is recursively enumerable in C by (2.2). Also, any finite set is certainly

recursively enumerable in C, so every set is

recursively enumerable in C. In particular, C"

00 00

is recursively enumerable in C, so C" < C1.

00 00 -M 00

2.4. Notation. Let f : A -+ B, g : C -+ E.

3

!y making the following diagram commute:

C<1 2 A '--) A $ C( C f+4 i +-Y 12

denote that y by f @ g If B and E are the same,

we also have a unique map y making

B

f +y g

A C--3A ( CE- C

1 i

2

commutative. Denote that y by f + g.

2.5. Remark.

R

does not admit arbitrary finite products.

Proof: Suppose P = N x N with projections p₁, P2 induced by 11 , ₂. Consider the following

diagram, where -3 is the constant function

1 if x Wo 0o if x = 00 ,x if x s K , and a(x) = W if x K or x = (x) = B c---+4B (0 EC E

CO N +px 00 P-N P₁ P₂ Claim 1: Proof: ker (y) = 0. If c # a e ker (y), a(a) = 1

#

C = p1 (CO) then = p (y(a)) *.

Claim 2: If a, b e R, then y(a) = y(b).

L+ Le t 0 :N -+P if x a₁ b if x = b if be defined

. Then 0 also makes the x = a

diagram commute, for pl(O(a))

6(b) = 1 = (a) and similarly

= p₁ (y(b)) =

p 1 ((b)) = 3(b). Also,

P2(O(a)) p2((b))

= p2 (y(b)) = a(b) = c = a(a)

= a(b). Then 6 = y by the

and similarly,

uniqueness of

so y(a) = 0(a) = y(b).

By claim 2, all values y(a) for a e K are equal; call the common value c.

00

Proof:

(x) (a)

14.

Claim 3: K = {x e Njy(x) = c}.

Proof: By claim 2, K C {x e Nly(x) = c}. If y(x) = c and x c K, then x = a(x) = p2(y(x))

p2 (c) = C *

Since y is induced by a partial recursive function, K = {x e Nly(x) = ci is recursively enumerable *.

00 00

Hence the product N x N cannot exist.

Some additional remarks might make clearer why

R,

cannot have products. The natural picture of the

00 00 C0

product N x N is {<x,y>ix, y e Ni, which would allow

pairs of the form <n,o> for example (n # co). We must have <n,00> l W, since pl(<n,oo>) = n, while

pl(w) = co. In our example above, we would expect, for

a number n E

K,

to have y(u) = <1,w>; but to define

such a y, we would need to be able to fill in the

second element of the pair, which we rould do only by

observing that n e K, an observation which cannot be

made effectively.

Proof: We show

R

has finite products. Let X, Y C N. The functions T, 7r₁, 7rr₂ are familiar. Let P = T"(X x Y). _{Then P is a product for X, Y,}

with projections 7 ₁ : P - X, r2 : P + Y. For let

f : Z -+ X, f₂ : Z - Y. The map z + T(f 1 (z), f2(z))

is the required unique map Z + P (uniqueness follows

from the fact that T is'l-1 and Onto).

2.7. Notation. Let f₁ : A -+ B ₁

both in

R2,

i.e. in

Rl

with ker 0. Y

f1 x f2 the unique map y : A x C -+ B

making the following diagram commute:

and f2 : C + E, Denote by x E with ker 0 A<-A x C - C + f2 f + B BK E E Pt B x E -?E

Let f₁ : A B 1 _{f: A + C with} ker 0. Denote by f @ g the unique map y:A + B x C with

ker 0 making the following commute: A p1 P2 B,,--- B x C - C 2.8. Remark. A x B = B x A; A @ B = B 4 A A x (B x C) = (A x B) x C; A $ (B $ C) = (A $ B) @ C We have seen that

R

admits finite but not infinite

coproducts nor does it admit finite products. However,

R2

does admit finite products. We see finally that:

2.9. Remark.

R2

does not admit infinite products (similarly for

_R

₁

_).

Proof: Let Y = N for r = 0, 1, 2,..., and

suppose P = X Y = X N, with projections p i P + N.

i i

Let f₀ : N +Y = N be the function which is every-where 0. Let f: N -+ Y = N (i > 0) be given by

1'

- if $ (x) _{fails to converge in < i steps}

O).(x)

1 if $(x) _{converges in < i steps.}

Note each f is total, so is a map in R1. Then

IT:N+P such that p y = fi for &ll i. Note y is total,

since it too is a map in

R',

Claim 1. If x,y c X, then y(x) = y(y).

Proof. _{This follows from the uniqueness of y,} as in the proof of (2.5). Let c be the common value

of y(x) (x C K).

Claim 2: K = {xjy(x) = c} which is recursively

enumerable * . This claim also follows as in (2.5).

Mitchell (1.17.3) proves the following useful

2.10* Theorem. _{The following statements are equivalent} in a category

_A.

(a)

A

_{has finite intersections and finite products} (b)

A

_{has equalizers and finite products}

18.

2.11. Axiom.

R

has equilizers.

Proof: Let a, : A + B. Let

C = {x e Ala(x) = a(x)}, and i the natural inclusion i : C A. Clearly i serves as Equalizer (c,3).

2.12. * Corollary.

R

2 has equalizers

Proof: Given a, : A -+ B in R₂. Let i : C4C-A

be the equalizer in _{R 1}

.

Then i is a monomorphism,

hence has ker 0, hence is an R2-map. Let f : E -+ A

in R2 such that af = $f. Then f factors through i, in

Ri,

say f = is. Since f has ker 0, so does s

(cf. 2.19 (11)), i.e. s is an R2-map.

2.13. * Corollary.

_R2

_{has pullbacks and finite} intersections.

We can actually do a little better than this.

2.14. Axiom. Let u : A -+ C, v : B -+ C be

R

1-maps with ker 0. Then in

R,

there exists a pullback P, P2 P -- 9 B + + A- C u

-L j. For x E A, y E B, u(x) 00 and v(y) W0 both $ w since p = {T(x,y)Ix e

U, v have ker 0. Let A,

00

y e B & u(x) = v(y)} {0}. Let

00

be induced by (i = 1, 2). Then the above diagram commutes.

Suppose _{u and v are induced by} 0 n

resp., and let f₁ : E - A f₂ : E - B be induced by

so uf_{1 = vf} 2. Define

e

as follows: dovetail computations $m

x).

If either is e(x) = T($k W, Clearly 6

4 ,(x))

convergent, let when convergent.

is the required map for the pullback diagram: f₂

3'

-1 Plj _tv - A C u Proof: _are pi resp. and _m given _for Onok(x) and W i xI

LU.

2.15. * Corollary.

R,

has finite intersections. Proof: _{An intersection of two monomorphisms is} just a pullback. Higher order finite intersections are gotten by induction.

2.16. * Corollary. If f : A -+ B is an

R

1-map with ker 0, then f has inverse images;

R2

has inverse images.

2.17. * Corollary.

R

₂ is finitely complete.

Proof: _{As Mitchell (II 2.6) points out} being-finitely complete is equivalent to having finite intersections and finite products.

We close this section with some simple remarks

about kernels and 0.

2.18. * Remarks. (i) Let

A

be an arbitrary category with 0. _{If u : A C-0O is a monomorphism,}

then u : A + 0.

u v

kernels, and A + C + B.

(ii) Let

A

have 0 and

21.

has ker 0; also ker (vu) = 0 => ker (u) = 0.

(iii) _{Let u K C+A,}

v : A4*'B, a : B + C such that vu = ker (a). Then

u = ker (av).

(iv) Let u K C!A be the kernel of d: A +E and say E

:E-+C has kernel 0.

Then u = ker (0a).

Proof: (i) Let v : 04_{4 A be the unique}

(mono)-morphism from 0 to A. Then uv : 0 - 0,

hence uv = 10. But then uvu = u, and u is a

monomorphism, so vu = 1 , hence u is an isomorphism. (ii) Consider: 0 + 0 + 0. The squares

A +C +B u v

are pullbacks, since u, v have kernel 0 (MItchell, p. 14), and 0 -+ B is a monomorphismn, so the outer rectangle is

a pullback (Mitchell, I 7.2), hence vu has ker 0.

For the second assertion, we have

ker cu) '.pker (vu) = _{0 as in Mitchell, p. 15, so by} part (i), _{ker (u)} = 0.

22.

K +0

A v

B +C

(iii) We have a pullback

The claim is that _{K 0 is also a}

u

AC-B +* C

v a

pullback. It is obviously commutative. Given a

map f : E - A such that avf = 0, we get a

unique map y : E -+ K such that vuy = vf. Since v is a monomorphism, uy = f. That such a y with the property that uy = f is unique follows from the fact that u is a monomorphism.

(iv) _{Consider the following diagram.}

K +0 +0

uf + +

A +E +C

a _S

Each square is a pullback, since 0 = ker (S) and u = ker (a) and 0 -+ _{C is a monomorphism, so the}

3. A Question of Balance

Recall that a category is balanced iff every

mono-epimorphism is an isomorphism. It is this property we wish now to explore for Rl.

3.1. Theorem. Let u : A'(-+B be a mono-epimorphism.

u is an isomorphism iff u is induced by a partial

recursive e which is 1-1 on its domain.

Proof: => Let u be induced by n' u~ by

$m. Define 0 as follows: given x, compute $ *(x) If it converges, compute $mn x). If this is convergent, see if $ n (x) = x. If so, give output $ n(x; if

not diverge.

Claim 1: 6 is 1-1 on its domain. For let

x, y e domain (0), with $n(x) = O(x) =

6(y)

= o (y). Then x o otp (x) =

4

o Y) = yo _{The x=}

m n M n*

Claim 2: 0 induces u, since if x e A, 0(x) is convergent and equals n Cx) = u(x).

: Let , induce u, with 0 1-1 on its domain4

given x, dovetail computations for e(0), e(1), ... until some convergence to x is found; if 6 (y) = x

is the first such convergence found, let 4(x) = y.

Note that G 1-1 =>

3

at most one y such that

0(y) = x. Further, if x e B, such a y exists and

00

belongs to A, hence 4 induces v : B + A. Clearly

00

v serves as inverse for u.

We shall see shortly that all infinite recursively

00

enumerable sets X are isomorphic in

R',

and that any set isomorphic to a recursively enumerable set is

itself recursively enumerable; we may use this to

3.3. Remark.

3

_{non-isomorphic objects A, B,}

and a mono-epimorphism u : A B. In particular,

u is not an isomorphism, so ₁ is not balanced.

Proof: It suffices to show

3

non-isomorphic sets X, Y _ N, and a mono-epimorphism u : Xc+"Y in

R1.

Suppose not. Let 0 be the following (total) recursive function: O(x) = least prime factor of x. Note 0 : N-.-P = {primes}, and P is recursively

set of integers with the properties that

U) 0 is 1-1 on Z, i.e. no two distinct elements of Z have the same least prime factor

and (ii) for each prime p,

3

z s Z such that p is the least prime factor of z.

Then 6 induces ZIC*OP, so Z

=

P, i.e. Z is recursively enumerable. However, there are clearly uncountably many such Z, and only countably many

recursively enumerable sets. *

In particular, let {p1, p2, ... } enumerate the

primes, and let {a,, a ... } be a (non-effective) enumeration of K. Let Z = {pnanln e N}. Then 0 induces ZC'P, but Z is not recursively enumerable, since Z recursively enumerable => K recursively enumerable clearly.

In fact, this result can be strengthened to show that

R

and

R2

are extremely unbalanced: given any infinite set X, I a set Y and mono-epimorphism

f : Y -+ X which is not an isomorphism. We need first

the following:

26.

3.4. Lemma. For every pair of infinite sets

X, Z, 3 a set Y and mono-epimorphisms

f Y4**X, g : YC-*?z.

Proof: Enumerate X = {x₁, x2, .. 1 and Z {Z 2 ... } (non-effectively). Let

Y = {<Xi, z >|i E NI and take f = 1st projection,

g

= 2nd projection.

3.5. Theorem. _{Given any infinite set X,}

3

a set Y and mono-epimorphism f : Yc+)X which is

not an isomorphism.

Proof: Since the number of partial-recursive functions is countable, and the number of infinite sets is uncountable,

3

a set Z such that there is no mono-epimorphism XC+iZ. Pick Y, f, g as in the Lemma: f : YC-*-X, g : Yc-+*Z. Note f is not an

isomorphism, for if so, we have gf 1 : XC+OZ *. The imbalance of

R

and R2 is disappointing, since it immediately forechoses the possibilities of normality and exactness, notions which have been of tremendous value in categorical studies of algebra.

27 .

Evidently

R

₁ is not an algebraic beast, but, as we shall see, there remains much of interest to say about

it.

Theorem (3.2) yields easily as a dorollary the fact that if A is recursively enumerable and

00

u : A - B is a mono-epimorphism, then u is an isomorphism. We shall, in fact, prove this result

categorically later on; at this point it interests

us as it raises the follo*ing question: does this property categorically characterize the recursively

enumerable sets, i.e. if A is such that every

mono-epimorphism with domain A is an isomorphism, must A be recursively enumerable? Our question is

00

answered in the negative by the following

3.6. Theorem.

3

a set _{X such that X,}

are immune (in particular, Xis not recursively enumerable) with the property that any mono-epimorphism

00

u : XC4OB is an isomorphism.

Proof: We construct X by stages. At each stage

at most a finite number of integers will be placed in

cd.

respectively.

Stage 2n: Consider Wn, the nth recursively enumerable set. If it is infinite, find integers an, bn e Wn different from all integers so far marked. Mark an with a 1 and bn with a 0.

If Wn is finite, go to the next stage.

Stage 2n+l: Consider $n. Let

Y2n+1 = {integers so far marked with a 1},

Z2n+l = {integers so far marked with a 0}. See

(not effectively) if $n is 1-1 on its domain. If so, go to the next stage; if not, there are 2 cases:

case (i). x, y e dom (4 ) such that

$n(x) = 4n(y) and both of x, y e Z2n+l. Then pick

such a pair and mark x, y with a 1 (if they are not already so marked), and proceed to the next stage.

case (ii) nCx) = n (y) (both convergent)

=>

either

x e Z2n+1 or y c Z2n+i. Then go immediately to the

29.

Let X = U Y 2n+1= {integers marked with a 1}. neN

Clearly the even numbered steps insure that X, X are both immune.

00

Now let _{u : XC**B be given, a mono-epimorphism,} induced by say $n. We want u to be an isomorphism, and by (3.1) it suffices to show that u is induced by some partial recursive 6 which is 1-1 on its

domain. If 4n is 1-1 on its domain, then u is an isomorphism and we are done. So suppose n is not 1-1 on its domain. There are two cases.

Case (i) x, y e dom ( n) such that n Cx) n(y) and both x, y i Z 2n+l* Then for some such pair ,

x, y s X, so $n is not 1-1 on X *.

Case (ii). $n Cx) = n (y) => either x e Z2n+l or y e Z 2n+l* Now Z 2n+ is finite, hence recursive,

so the function

e

is partial recursive, where rn(x) if x i Z 2n+

0(x) =. Note

divergent if x e Z2n+l

domain 6. Since X Cdom ($n) (u has kernel 0) and X

fl

Z₂n+₁ = $, X C dom 6, so 0 induces u. Finally 0 is ll on it5 domain, for if not, we would be in case (i). Hence n is an isomorphism.

4. _{Image, Coimages, Reversible Monomorphisms,} Quasi-Unions and Essential Domains

A natural question to ask of a category is whether arbitrary maps have images. Recall that the image of

a map a : A -+ B is a monomorphism i : I4 B such

that a factors through i, and such that for any other monomorphism

j

: JC-vB such that a factors

through j, i factors through j.

4.l. Remark In

R

it is not in general the case that the natural inclusion of the set theoretic image serves as the categorical image.

Proof: Let f : Z'4-P be as in (3.3), with P

00

recursively enumerable but Z not recursively enumerable. 00

Just suppose 1p : P + P serves as Im (f), Since f factors through f, 1 must also factor through f.

U b

A moment's thought establishes that this forces f to be an isomorphism *. It should be pointed out that f does have an image, namely itself.

The question of general images in

R

remains unsettled. The following is known, however.

4.2. Proposition. Let A be a recursively enumerable set, and let f : A + B. Then f has an

image; in fact the natural inclusion of the set theoretic image works in this case.

Proof: Let _{f"A be the set-theoretic image of f,} and let i : f"A -+ B be the natural inclusion. The

assertion is that i = Im (f). Clearly f factors through i, say f = i7. Let j : J B be a

A A

monomorphism such that f = Jf for some f : A -+ J.

We must show i factors through j. Let $n induce f; since A is recursively enumerable we may assume

00

dom (n) C A. Define a partial recursive function 0

00

as follows: given x. begin a Stand'ard dovetailing

32.

t he first integer such that this procedure gives

n (y) = x, if any such exists. Let O(x) = f(y). Claim 1: 6 induces k : f"A -+ J,, For let

x e (f"tA).. Then y e A such that n(y) =(y) X.

Then f(y) s J.

Claim 2: jk = i. For if x c (f"A) , then k(x) = f(y) for some y c A such that n(y) = X.

A

Then jk(x) = jf(y) = f(y) = iT(y) = i4)n(y) = i(x).

Fortunately,

R

₁ behaves much more nicely with

respect to coimages. In fact, if a : A -+ B is a map,

then the induced map a : A++cat"A serves as the coimage, and we have the following

4.3. Axiom. Every

R

₁ map a : A + B factors

- i

into Aa ++ B, with A ++ I the coimage of a.

Proof: Let : A -+a"A be induced by a, and i the natural inclusion. Clearly a = i. Let

: A + J be an epimorphism such that for some

6-) C

x F J,

3

y c A such that $(y) = x. Then

k(x) = kS(y) = a(y) c a"A, so k factors through i. We have the following diagram

A -I

I,

with iU = k. monomorphism, commutative.

Then i3= a=k so a = j , so the = ij, but i is a entire diagram is 4.4. Definition. Let u : u A - B u C i A + B be be the factorization a monomorphism, through the

coimage. Note U is a mono-epimorphism. u is reversible iff u is an isomorphism.

4.5. Remark. u : ACa;B is reversible iff u

induces an isomorphism A + u"A.

4.6. * Remark. Let a : A -+ B have coimage

factorization A DB

C

. Then v is reversible. v

Proof: Let v have coimage factorization

V C C-"-'2B

. We show v a: A + E also serves

as coimage (a), hence that v is an isomorphism by

the universal property of coimage. Clearly v a is

an epimorphism, being the composition of epimorphisms, a

and a factors through v a . Let A -+ B be given.

Then j: G -+ C such that jf = a. Then

Vjf : G + 7 and vjf = v a.

4.7. * Corollary. Equalizers (hence kernels) are

reversible.

Proof: Let a,B : A -+ B, and let i : E A be

EqU (a, f). _{Factor i through its coimage i =}

_jf,

with

j

: E + A a monomorphism. Claim : j = Equ (a,$). We have aji = ai = 0i = aJr, but is an epimorphism,

so aj =j. Given f : C + A such that _af f, we

have g : C E such that ig = f. Then g =C and jig ig = f.

We seek a concept of union in Fl, but it turns out to be more convenient to define a related concept,

which is simply the notion of union restricted to reversible monomorphisms. To be more precise:

4.8. Definition. Let {u i : A i-+ A} be a

family of (not necessarily reversible) monomorphisms. A quasi-union u : A' C+ A is a reversible monomorphism

such that

(i) each u factors through u

(ii) given any object B, map f : A -+ B, and reversible monomorphism v : B' C B, if each ui gets

carried into v, then so does u.

4.₉ . * Theorem. Quasi-unions are unique (if they exist) up to isomorphism as subobjects.

Proof: Let ul : A C+ A and u" : A" 4 A be two quasi-unions of {u ::A 4 Al. Since u" is reversible, and 1_{A carries each u}

into u", u' is also carried into u". We have

'35.-w

A' + A"

u U" and since lAu is a monomorphism, so

A - A 1 A

is w. Hence ul precedes u" as subobject _{of A;} similarly u" precedes u', so the two are isomorphic

as subobjects of A. (cf. Mitchell, p. 6).

4.10. Axiom.

R

admits arbitrary quasi-unions.

Proof: Let {u i: A C A} be given. Then each u i"A _{C A,}

the natural

so A' = U ui"A C A. Let u -A' + A be

is

inclusion so u is reversible and every factors through u.

Let f : A -+ B _{be given and let v :} B' C+. B be

a reversible monomorphism _WOLOG, v is the natural

inclusion (otherwise we may replace B' by the

isomorphic subobject v"B' C B). Suppose f carried

each _u. into v. Then fu "A C B' for each i, so f"A' = f"(Uu i"Ai) = U fu "A CB', so f carries

i i

37.

Since no mention shall be made of traditional unions, we shall reserve the notation U u for the

i quasi-union.

The following fact is worth noting. Since no further use is made of it, it is not presented as an axiom.

4.11. Fact. Let {ui : A iC A} be a family of reversible monomorphisms. Then u exists.

Proof: WOLOG, assume each ui is a literal inclusion. Let A' = A and v : A'(-'l A the

natural inclusion. Then v factors through each of the ui. Let w : B -+ _{A factor through each u.;} then for each i, w"B C A , so w"B n A = A', so w factors through v.

4.12. Definition. Let a : A + B be an

Rl-map.

Let D be a singleton (i.e. an R2-null object).

Let C = {u : D-+ Ajau P 0}. _{The essential domain of a,} denoted Edom (a), is U u.

4.13. Remark. Let a : A -+ B. The natural

inclusion {a e Aja(a) convergent}w _C A serves as Edom (a).

5. Singletons

Clearly singletons will play a very important role in this work. The present section is devoted to

exploring some of their basic categorical properties. (Throughout this section, let D denote a singleton).

5.1. Axiom. Every map in

R,

with D as domain

and kernel 0 is a monomorphism.

5.2. Axiom. B # 0 iff

3

a monomorphism

v : D C+ B.

5.3. * Remark. Let u : A + D be a monomorphism. Then A = 0 or u is an isomorphism.

Proof: If A # 0,

3

v : D + A a monomorphism. Since uv : D -+ D is a monomorphism, it has ker _0,

hence uv = D _{m iap D + D). Then uvu}

= lDu = u = ulA and u is a monomorphism, so vu = 1_{A* Thus u is an}

39.

5.4. * Remark. If A

$

0, and a : A - D has

Edom (a) # 0, then a is an epimorphism.

Proof: Suppose we have g, h : D + E such that

ga = ha; we must show g = h. Let w : BC+ A be Edom (a).

Since _{B p 0,}

3f:

DC4 B. Note awf : D-+D is not 0;

it follows that awf has kernel 0 (if not then

A q : D ( D such that awfq = 0, but q must be 1D since it is a kernel 0 map from D to D). Further awf = 1D. Then g = glD = gawf = hawf = hlD = h.

5.5. * Remark. Let A (# 0) be such that every

map with A as (categorical) domain and kernel 0 is

a monomorphism. Then A is a singleton.

Proof: Let v : A -+ D with kernel 0 (such exists

since D is R₂ null). By hypothesis, v is a

monomorphism, so by (5.3), A 0 or v : A + D. 5.6. * Remark. D x D D.

Proof: It suffices to show that D x D is R2 null.

Given object A, we have v : A -+ D _{with kernel 0,}

U .

a, : A + D x D with kernel 0. Let p₁, p₂ be the projections. Then p1a, piS : A -+ D with kernel 0,

hence pla = plO. Similarly, p₂a = P2r. Hence a =

The following axiom asserts that every object is made up of singletons:

5.7. Axiom. For every object A, 1A = U{u : D-+, Al Proof: Clearly 1A is reversible, and each

u factors through 1 Given f : A - B, and v : B' C+ B, reversible such that each u gets

carried into v, we may sssume that v is a literal

inclusion. Then each u"D CB', and since each u i DC* A picks out a member of A and conversely, it follows that A

C

B', so 1A gets carried into v.

As we shall see subsequently, the following axiom asserts that eacy singleton is recursive.

5.8. Axiom. Let u : D C+ B. Then

I

E and

a : B - E such that u = ker (a).

Proof: Let D = {O,oo}, and let x = u(O) e B. 00

Define 0(y) =

y if y d x

convergent if = x.

Then 0 induces

B - N with kernel u clearly.

Actually we shall want another formulation of this recursiveness, to wit:

5.9. Axiom. Let u : DCE*B be given. Then

3

w : A C-0B such that B = D $ A, with injections u and w.

Proof: Simply take the natural inclusion

B - u"Dc4B. 00

6. Complements

Throughout this section, D denotes a fixed singleton.

6.1. Definition. Maps u : A + C and v : B + C are disjoint iffm monomorphism DeN9C factors through both of them. Clearly u and v are disjoint

iff (u"A)₀₀ and (v"A)₀ are disjoint sets.

6.2. * Remark. Let u : A -+ C ahd v : B +

42.

Then au and av are disjoint.

Proof: Suppose w : D -+ E factors through both au and av, i.e. we have commutative diagrams

u a A + CC E V a B - C'+ E and g D D w

Then auf = w = avg, and a is a monomorphism, so uf = vg. Moreover, since w has kernel 0, so does uf = vg : D -+ C. _{Since D is a singleton, uf = vg}

must be a monomorphism (5.1) D4+ C factoring through both u and v *.

6.3. Definition. Maps u : A + C and v : B + C

are complementary iff

(i) they are disjoint and

(ii) every monomorphism D 4 _{C factors}

through one of u or v.

43.

(u"A). and (v"B)O are relative complements in C.

00

Given a map a : A -+ B, a complement of a is a

reversible monomorphism C 4 B, complementary to a.

6.4. * Theorem. Complements (when they exist) are unique, up to isomorphism as subobjects.

Proof: Let

8

: C -+ B and

8'

: C' 4 B be

two complements to a. Let v : D + C. Then av

factors through 8, hence not through a, hence through

8'. We have the following situation:

D -- -- >Ct

4?

, i.e. a carries v into

8'.

C

c-

V

Since 1C = U{v : D - C1, it follows that # carries 1C into 8', hence that a precedes 8'. Similarly

8' precedes 8, so the two are isomorphic subobjects

44.

6.5. Axiom.

R,

admits complements.1

Proof: Given a : A -+ B, the natural inclusion

i (B - a"A)"f4 B serves as the complement.

We shall denote the complement of a : A -+ B by

ca : K + B.

7. The ordering _<c

7.1. Remark. Let X N. Then in

R

X X join $ via Xx[2x1.

7.2. Lemma. Suppose in

R',

X = Y join V for

some V C N, and Y = X join W for some W CN. Then X = Y.

Proof: If X and Y are finite, it is easy to

see they must have the same cardinality, hence are isomorphic. Assume for the duration that X and Y

are infinite. _{Let $n} induce the map X join W + Y, 1 This axiom is actually superfluous. In fact, with the aid of (III, 1.1), one can easily show categorically that

for any map a : A B U{u : D 4 Blu does not factor through a) serves as the complement.

45.

with inverse induced by $n,. Let $m induce the map

Y join V + X, with inverse induced by $m,. The maps

Y 4 Y join U and Xc+ X join W are induced by Xx[2x].

Define 01, 02 as follows:

On , o $ (2u)

e (u) = n' 2 (note divergent if 0 $~a0 4n (2u) is odd)

0m rft (2u)

0,2(u) = 4)'~ 2

Now let R be a recursively enumerable set of ordered triples of integers, gotten from the following instructions: list (a, b, c) iff

(i) a = 6₁(u) for some u & $ n n' '1 n (2u)

is convergent and = 4)n (2u) = b & c = 1

or (ii) b = 2 (v) for some v & 4m 0 m ' c m (2v)

is convergent and = $m (2v) = a & c = 2.

By dovetailing computations, one can effectively generate R in some order. Let

46. R = {(al, b₁ , c= 1 or 2. Claim 0: similarly, Claim 1: as an aj in c 1 ), (a2, b2, c2), ... }, where u e X => v e Y => For each

61(u) = u & no n On(2u) =

%n

(2u) e Y

0₂(V) = v & 0mO mi m(2v) = $m(2v) c X

U e X, u appears at most twice

R; one time for sure with ci = 1, and possibly another time with ci = 2. Analogously for

each V E Y.

Proof: Let u e X. Clearly the triple

(U = 0₁ (u), $m(2u), 1) e R. Suppose (u, b, 1) e R. Then u = 1 (t) for some t such that

On(2t) = n 0 0n' o n (2t) and b = $n (2t). Now

U n' On(2t)

u = 2 _{, so}

2u = 0n' A n (2t), so

On (2u) = $n 0 0 n' a n(2t) = $ n(2t) = b. Thus

(u, $n(2u), 1) is the only triple in R with first coordinate u and last coordinate 1.

4'.

Suppose a triple (u, b, 2) e R. Then b = 0₂(v) for some v such that

m m' C m(2v) = 4m(2v) = u. Then $m,

4(2v)

b = ₂ so, since u e X, $ (u) b = 2-7- C

4,(u) _{. Note 4,m,(u) must be even} $m,(u) C Y join U, and is even, hence

Claim 2: For each (a, b, c) Proof: Immediate

Notation: For x

s R, a e X <=> b s Y. from claims 0, 1, and their proofs. s X, let fi(x) _{= unique y}

such that (x, y, 1) s R. For v e Y, let

f₂(v) = the unique u such that (u, v, 2) e R.

We now list a set R' of ordered quadruples of

integers inductively as follows. Say

(a t, b1', c1t, d 11), ... (ak', bkt, ck', dk') have

been listed, subject to the following conditions: If c ' = 1, then (d.', bit, 2) E R; if c

j

= 2,

to the following procedure.

(1) For u not yet appearing as some a ,

search in R for u = a with c = 1. If such is found, momentarily stop dovetailing, and define f(a ) as

follows: If b 1 b for all i < k with c ' = 2,

let f(a ) =b . Say b =b c 1' = 2. If

(i < k) and

d t b ' for all k < k such that C = 2, let f(a4) = di '. If d 1' = b 2' for some i2 < ₂ k with c 2 = 2, continue the procedure

1i₂

with d2' _{. In this way the sequence}

b , d 1' , d 2

is built up. Let f(a ) be the first repetition in the sequence if such occurs, otherwise f(a ) = d i

s for least s such that d b for all k < k with c zI = 2, We shall say a is "searching for"

f(a ). Now go back over all computations that have so far converged according to this procedure,

including part (2) below (only finitely many have thus

far converged, since each such computation must arise

from a number preceding a. in the list of tuples

in R), and see if any of them are searching for a j If v is searching for aj, and is the first such,

then add the quadruple (aj, v, 2, b ) to the list. Note that it satisfies the conditions. If no

computation is presently searching for aj, continue the dovetailing computation by searching through R

for some t such that (t,f(a ), 2) e R. If one is found either in the course of this computation, or subsequently in the course of some other computation, add the tuple (aj, f(a ), 1, t) to the list. Note that it satisfies the condition.

(2) Simultaneously, perform analogous computations for all v which have already appeared as some a i

but have not yet appeared as some b '. In the course of this computation we shall say v = b is searching

for g(b ). At any given stage in the process there

will be at most one tuple appropriate for adding

to the list. If there is one such, add it and begin

recursively enumerable.

Claim 1*: If (u, v, c, d) e R', then

u C X <=> v C Y.

Proof: Say (u, v, c, d) c R' and c 1. We must be in situation (1), i.e.

(u, v, c, d) = (aj, f(a ), 1, t). Note

(a4 , b , c ) c R, so u e X <=> b e Y <=> b ' = b c Y <=> a ' 1 X <=> d ' e Y <=> b ' Y <=> 1 12 a ' c X <=> d 1 t Y <=> ... <=> v = f(a ) = d i I Y. 2 i2 s

If c = 2, the argument is similar: here the tuple

must have form (g(b ), b , 1, t).

Claim 2*: (V u C X) (3 at most one v)

(3

c)(Jd)[(u,v,c,d)

c R'

analogously, ( v c Y) (

3

at most one u) (3 c)(3 d)[(u,v,c,d) e R'1

Proof: Supppose (u, v, C, d) c R' was put in as

the k+1st tuple, u e X, and (u, v', C', d) c R' was

put in earlier, and this is the 1 place of such

difficulty. If c = 1, we have an immediate contradiction,

51.

computation at stage k. So say c = 2. Then v = b and u = g(b ) for some J. Since u c X, v = b e Y. Subelaim: Since v e Y, the appropriate sequence

a = d ' , d 1' , d _{2 , ... contains no repetitions..}

Proof: of subclaim: We have (a , b , 2) e R, aj = a a =s1, so (aj, b' 1, d 1') e R', with

b b ( by const. of type (2)), hence

(di ' b 2) e R'. Note b e Y => a E X =>b ' e Y >

d CX. Since b bi ' and both

(aj, b , 2), (d 1', b 1', 2) c R, a # d i' by claim 1'.

Similarly, the sequence consists of all distinct members. There are now two cases:

(i) c' = 1. _{Since the above sequence has no} repetitions, u = g(b ) cannot equal a 1 for any

>2.

(ii) c' = 2.

We have sequences v = b , a = d 0, v, U 1, V2, u2 ***,vsus=t

and v' = bj,, aj, = di , v ₁ , u', v2', U2 v t',s = U,

satisfying:

(i) (aj1 V, 2), (u1 , v1 , 2), ...,(u=ussvS2) eR

(a '$ v', 2), .. , (u ' = u, v t1 2) e R;

and (ii)

have already been listed in R', as have been

(a ' v ' 1 , u I),...(ut, ', 1 , i u t')9 By

claim 1 and the fact that R' has no repetitions on each of the first two coordinates so far we conclude that

Us = u = Ut; v s Vt,; U = us-l ***

There are three cases:

(i) s = t, i.e. v' = v

(ii) s < t. Then v = v ' for some Z, and would already have appeared as a second coordinate in the list,

so could not now have been considered for a type (2) calculation.

(iii) s > t. Then v' = v for some X < s. Then (u2_1, vx,1 1, u5 ) already appears in R', but R'

contains no repetitions as yet, so u = u = u . It follows that v x ₁ = v5 by the "1-1 ness" of R

with respect to X, Y (Claim 1), which implies _{that R'} does have repetitions so far *

This includes the proof of half of claim 2 The other half is proved similarly.

*.

Claim 3*: _{( V u E X) (3 v, c, d)[(u, v, c, d) e R'];}

(V v e Y) ( _{3u, c, d) [(u, v, c, d) e R'].}

Proof: Let u e X. _{The preliminary search for} some j such that u = a with cj = 1 eventually succeeds, _{Suppose k tuples have been listed by such} time as a convergent computation is reached. One

immediately defines _{f(a ) and begins searching for it.} Eventually, since f(a ) E Y, _{a t such that}

54.

(t, f(a ), 2) e R will be found; however, by this time more tuples may have been added to R', say tuples k+l through k. Suppose u has still not been added as a first coordinate in any tuple.

Subclaim: _{At this point in the construction,}

if one defines f'(a ) _{to be the new sought integer,} fr(a') = f(a ).

Proof of subclaim: It suffices to show f(a ) b s for any s such that k+1<s<k . Suppose f(a) =b

for some such s. If c' = 1, then since b5 I= f(a ) e Y, as' e X and we have

(as, f(a ), s , e R'. Then we have sequences

as',

vl'

, u2 P

v

2, *.., u', f(a ) and

aj, v₁, u22 2, ... , up, f(a ), as in the proof of

the previous claim. _{As was shown there,}

u ' = up, u = UP-1,... Also, v1 = bm for some

m < k. There are three cases:

(i) u = aj = a * (u has not yet appeared as first coordinate)

55

(ii) as' = u. for some uj in the sequence

aj, v₁ u 2 , ... ; but all such uj actually appear as

first coordinttes before the k+lst tuple. *

(iii) a = ui I for some ui in the sequence as vt l, u₂ , ... ; but each u i' actually appears as

first coordinate before the sth tuple *,

Hence if f(a ) = bs for some s' such that k + 1 < s < X, then cs' = 2; but then in the

computation for bs' we would actually discover some t such that (t, bs' = f(a ), 2) e R. This would conclude our search for f(a ) in the computation for a,

and according to the instructions, the tuple (a , f(a ), 1, t) would have been added at this time. * (aj = u has not

yet appeared as first coordinate).

This establishes the subclaim. Since f'(a) =f(a

and the search for f(a ) has been concluded, the tuple (aj, f(a ), 1, t) is now added to R', establishing half the claim. _{The other half is established similarly.}

Putting claims 1*, 2*, and 3* together, we have:

(Vu e X)(J!v)(3c)(3d)[(u,v,c,d) e R'] and that

V e

Y

(V v s Y)(3!u)(3c)(Jd)[(u,v,c,d) e R'] and that u e X.

Clearly the following partial recursive function a induces an isomorphism X + Y in R2

':

a(u) = b ',

where b ' is the first integer such that c, d are discovered making (u, b ', c, d) e R', The inverse is induced by (v) = aj, where a is first integer such that

3

c, d making (aj v, c, d) e R'.

7.3. Remark. A = B in R, <=> A = B in

R

<=> A = B in R21'.

20 0 0

7.4. Definition. GIven A, B, objects in

Rl.

A is categorically smaller than B (A <c B) iff

ah object C such that B A $ C. Note c is well-defined on isomorphism classes.

7.5. Theorem. _-c< is a partial ordering. Proof:

(i) < is reflexive: _{Given A, A = A @ join $}

-c _{00 Go}

in R₂

',

so A= (A join

i)

= A e 0 in

Rl.

00

(1i) < is transitive: _{Suppose B = A e E, C B $ F.} -C

Then C = (A $ E) $ F = A e (E e F).

(iii) _-C < is antisymmetric: _{Suppose A} < B & B < A.

-c

-Then 3 C, E such that B = A $ C, A = B $ E. Then B = (A @ )00

=

_{A join C; A = B join E; so by the lemma,}

0600 00- 0 00

00

A B in R₂ ' , hence A B. 00 00

The ordering <c has been studied extensively by Dekker and Myhill. Letting fl = the set of isomorphism classes in

R

2

',

and viewing @ as a binary operation,

they show (2, @) _{is a commutative, partially ordered} groupoid with the finite refinement property. Our

work shall not take this direction, since it seems very unlikely that the < ordering defined in w-recursion

theory will tnt gewralb _{to a partial ordering in} generalized a-recursion theory for admissible a. Aside from one application in the next chapter, in

providing a second characterization of N, we shall make no further use of c'

56.

Chapter II

Recursion Theoretic Properties Expressed Categorically

This chapter aims to translate various basic recursion theoretic notions into statements about Surprisingly, many such notions are amenable to this procedure, including recursively enumerable, recursive,

immune, cohesive, index (G8del-numbering) and

productive. Throughout this chapter, D denotes a singleton.

1. Characterizations of

N

1.1. Definition: An object A in

R

_{1 is finite} iff it is a finite coproduct of singletons. If A is not finite or 0, A is infinite.

1.2. Remark. A is finite iff A is finite.

00 In fact, A= q i=l,. . D iff A = n. o

1.3. Lemma. X C-N is infinite recursively

enumerable iff _{N = X in R2}t

_.

Proof: => : Say X -is infinite recursively enumerable. Then X is the set-theoretic image of a