every closed category is enriched over itself
sergei-...@rambler.ru
In one paper on category theory I found the following result: every
closed category K is automatically enriched over K. I wonder how this
is proved.
Thank you in advance,
Sergei Akbarov
Philippe Gaucher
> Dear colleagues,
>
> In one paper on category theory I found the following result: every
> closed category K is automatically enriched over K. I wonder how this
> is proved.
Probably you mean a cartesian closed category C ? This means that the
functor -xY (product by Y) has a right adjoint. i.e. there is a
natural bijection of sets C(XxY,Z)=C(X,Hom(Y,Z)) where Hom(Y,Z) is an
object of C. You can replace each homset C(X,Y) by the object
Hom(X,Y). The composition is defined as follows.
1) Consider the bijection C(Hom(X,Y),Hom(X,Y))=C(XxHom(X,Y),Y). The
image of the identity of Hom(X,Y) gives a morphism XxHom(X,Y)-->Y
called the evaluation map.
2) The evaluation map gives XxHom(X,Y)xHom(Y,Z)-->YxHom(Y,Z)-->Z hence
an element of :
C(XxHom(X,Y)xHom(Y,Z),Z) = C(Hom(X,Y)xHom(Y,Z),Hom(X,Z)),
that is a morphism Hom(X,Y)xHom(Y,Z)-->Hom(X,Z). It's your
composition.
Try <http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html> and
<http://en.wikipedia.org/wiki/Cartesian_closed_category>.
pg.
sergei-...@rambler.ru
Thank you very much for your answer! Actually I had in mind closed
monoidal category (not necessarily cartesian) - and I forgot to say
that it should be symmetric also.
I understand the definition of morphism Hom(X,Y)xHom(Y,Z)-->Hom(X,Z),
since
it is defined also in the paper I mentioned above (this is a paper by
E.G.Shulgeifer in russian hand-book "General algebra"). But a puzzle
for me is why the corresponding diagrams (for Hom's) must be
commutative. I would appreciate very much, if somebody explains me
this...
Thank you again,
Sergei
Philippe Gaucher
> Actually I had in mind closed monoidal category (not necessarily
> cartesian)
Of course, monoidal works as well.
I believe that your theorem is proved in
<http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html> p15.
pg.
sergei-...@rambler.ru
I am a specialist in Functional Analysis, not in Category Theory, and
I must say that I do not understand Kelly's hints on page 15 of his
famous paper:
"Yet direct verifications, although somewhat tedious, are nevertheless
fairly straightforward if the order below is followed."
"Verification of the axioms (1.3) and (1.4) is easy when we recall
that, because of the
relation of e to the of (1.23), the definition (1.28) of M is
equivalent to e(M 1) =
e(1 e)a."
Not long ago I tried to reconstruct another "evident proof" in this
science, after a number of futile attempts to find somebody who could
explain me this, and finally my own proof took 5 pages with 23
diagrams. In my opinion, this way of writing mathematical texts
borders upon indecency.
I believe, there must be something more intelligible on this theme... Do
you know a textbook (or a paper), where those things are described in
details?
Sergei
Philippe Gaucher
> Not long ago I tried to reconstruct another "evident proof" in this
> science, after a number of futile attempts to find somebody who
> could explain me this, and finally my own proof took 5 pages with 23
> diagrams. In my opinion, this way of writing mathematical texts
> borders upon indecency.
Maybe try the Handbook of Categorical Algebra by F. Borceux (vol. 1 I
think -there are three volumes). But the verification is very
simple. You have to use the bijection coming from the adjunction. And
the coherence axioms do the job. For the associativity, you have to
prove that the two maps
(C(M,N)xC(N,P))xC(P,Q)-->C(M,P)xC(P,Q)-->C(M,Q) and
C(M,N)x(C(N,P)xC(P,Q))-->C(M,N)xC(N,Q)-->C(M,Q) correspond to the same
map MxC(M,N)xC(N,P)xC(P,Q)-->Q.
So you have to compare:
Mx((C(M,N)xC(N,P))xC(P,Q))-->(Mx(C(M,N)xC(N,P)))xC(P,Q)-->(NxC(N,P))xC(P,Q)-->PxC(P,Q)-->Q
Mx(C(M,N)x(C(N,P)xC(P,Q)))-->(MxC(M,N))x(C(N,P)xC(P,Q))-->Nx(C(N,P)xC(P,Q))-->PxC(P,Q)-->Q
The monoidal structure x is associative up to isomorphism and the
first coherence axiome of p7 of Kelly's book completes the proof of
the associativity. For reference about coherence for associativity,
see "Categories for the working mathematician" by Mac-Lane p160.
If -really- you want more details: The first square
Mx((C(M,N)xC(N,P))xC(P,Q))-->(Mx(C(M,N)xC(N,P)))xC(P,Q)
Mx(C(M,N)x(C(N,P)xC(P,Q)))-->(MxC(M,N))x(C(N,P)xC(P,Q))
is commutative by the Pentagone axiom.
The second square
(Mx(C(M,N)xC(N,P)))xC(P,Q)-->(NxC(N,P))xC(P,Q)
(MxC(M,N))x(C(N,P)xC(P,Q))-->Nx(C(N,P)xC(P,Q))
can be decomposed on
(Mx(C(M,N)xC(N,P)))xC(P,Q)-->((MxC(M,N))xC(N,P))xC(P,Q)-->(NxC(N,P))xC(P,Q)
(MxC(M,N))x(C(N,P)xC(P,Q))===(MxC(M,N))x(C(N,P)xC(P,Q))-->Nx(C(N,P)xC(P,Q))
The left-hand square is commutative by the Pentagone axiom. And the
right-hand square by the natural transformation
(-xC(N,P))xC(P,Q)-->-x(C(N,P)xC(P,Q)) with - replaced by MxC(M,N)-->N.
The third square
(NxC(N,P))xC(P,Q)-->PxC(P,Q)
Nx(C(N,P)xC(P,Q))-->PxC(P,Q)
can be decomposed on
(NxC(N,P))xC(P,Q)===(NxC(N,P))xC(P,Q)-->PxC(P,Q)
Nx(C(N,P)xC(P,Q))-->(NxC(N,P))xC(P,Q)-->PxC(P,Q)
The left-hand square is commutative by the Pentagone axiom. The
commutativity of the right-hand square is obvious.
The square
PxC(P,Q)-->Q
PxC(P,Q)-->Q
is commutative.
Conclusion : this proof is totally useless and mechanical.
pg.
sergei-...@rambler.ru
First, as I understand, by C(M,N) you mean Hom(M,N). So your first
steps should be written as follows:
> For the associativity, you have to prove that the two maps
> (Hom(M,N)x Hom(N,P))x Hom(P,Q)--> Hom(M,P)x Hom(P,Q)--> Hom(M,Q) and
> Hom(M,N)x( Hom(N,P)x Hom(P,Q))--> Hom(M,N)x Hom(N,Q)--> Hom(M,Q) correspond to the
> same map Mx Hom(M,N)x Hom(N,P)x Hom(P,Q)-->Q.
> So you have to compare:
> Mx(( Hom(M,N)x Hom(N,P))x Hom(P,Q))-->(Mx( Hom(M,N)x Hom(N,P)))x Hom(P,Q)-->(NxHom(N,P))x Hom(P,Q)-->Px Hom(P,Q)-->Q
> Mx( Hom(M,N)x( Hom(N,P)x Hom(P,Q)))-->(Mx Hom(M,N))x( Hom(N,P)x Hom(P,Q))-->Nx(Hom(N,P)x Hom(P,Q))-->Px Hom(P,Q)-->Q
And, second, I don't agree with the last line - I don't see why this:
Hom(M,N)x( Hom(N,P)x Hom(P,Q))--> Hom(M,N)x Hom(N,Q)--> Hom(M,Q)
- must be equivalent to this:
Mx( Hom(M,N)x( Hom(N,P)x Hom(P,Q)))-->(Mx Hom(M,N))x( Hom(N,P)x
Hom(P,Q))-->Nx(Hom(N,P)x Hom(P,Q))-->Px Hom(P,Q)-->Q
Are you sure there is no carelessness here?
I did not find in our libraries the book you cite, by F.Borceux. Of
course, I'll try to search anywhere else...
Sergei
Philippe Gaucher
> Hom(M,N)x( Hom(N,P)x Hom(P,Q))--> Hom(M,N)x Hom(N,Q)--> Hom(M,Q)
>
> - must be equivalent to this:
>
> Mx( Hom(M,N)x( Hom(N,P)x Hom(P,Q)))-->(Mx Hom(M,N))x( Hom(N,P)x
> Hom(P,Q))-->Nx(Hom(N,P)x Hom(P,Q))-->Px Hom(P,Q)-->Q
>
> Are you sure there is no carelessness here?
With the coherence axioms, you can remove the parentheses. You then
have a commutative diagram:
MxHom(M,N)xHom(N,P)xHom(P,Q)-->MxHom(M,N)xHom(N,Q)
NxHom(N,P)xHom(P,Q) -->NxHom(N,Q)
PxHom(P,Q) -->Q
The passage from the first to the second line by the naturality of
*xHom(N,P)xHom(P,Q)-->*xHom(N,Q) applied to
*=(MxHom(M,N)-->N).
The map NxHom(N,P)xHom(P,Q)-->PxHom(P,Q)-->Q yields a map
Hom(N,P)xHom(P,Q)-->Hom(N,Q) by adjunction, hence a map
NxHom(N,P)xHom(P,Q)-->NxHom(N,Q) and the commutativity of the second
square.
You have the composite of two commutative squares which is a
"homotopy" between
MxHom(M,N)xHom(N,P)xHom(P,Q)-->MxHom(M,N)xHom(N,Q)-->NxHom(N,Q)-->Q
and
MxHom(M,N)xHom(N,P)xHom(P,Q)-->NxHom(N,P)xHom(P,Q)-->PxHom(P,Q)-->Q.
pg.
Philippe Gaucher
> NxHom(N,P)xHom(P,Q) -->NxHom(N,Q)
> PxHom(P,Q) -->Q
> The map NxHom(N,P)xHom(P,Q)-->PxHom(P,Q)-->Q yields a map
> Hom(N,P)xHom(P,Q)-->Hom(N,Q) by adjunction, hence a map
> NxHom(N,P)xHom(P,Q)-->NxHom(N,Q) and the commutativity of the second
> square.
This argument is maybe not enough detailed:
Consider the natural bijection C(*,Hom(N,Q))=C(Nx*,Q). Replace * by
the map Hom(N,P)xHom(P,Q)-->Hom(N,Q) corresponding to the composite
map NxHom(N,P)xHom(P,Q)-->PxHom(P,Q)-->Q by the adjunction. You obtain
a commutative square of sets
C(Hom(N,Q),Hom(N,Q)) ==== C(NxHom(N,Q),Q)
C(Hom(N,P)xHom(P,Q),Hom(N,Q))==== C(NxHom(N,P)xHom(P,Q),Q)
1) Start from the top-left with the identity of Hom(N,Q). The path
C(Hom(N,Q),Hom(N,Q))->C(NxHom(N,Q),Q)->C(Nx(Hom(N,P)xHom(P,Q)),Q)
gives you the composite NxHom(N,P)xHom(P,Q)->NxHom(N,Q)->Q.
2) Start from the top-left with the identity of Hom(N,Q). The path
C(Hom(N,Q),Hom(N,Q))->C(Hom(N,P)xHom(P,Q),Hom(N,Q))->C(NxHom(N,P)xHom(P,Q),Q)
gives you NxHom(N,P)xHom(P,Q)-->PxHom(P,Q)-->Q.
pg.
sergei-...@rambler.ru
As I understand, you complete your diagram
MxHom(M,N)xHom(N,P)xHom(P,Q)-->MxHom(M,N)xHom(N,Q)
NxHom(N,P)xHom(P,Q) -->NxHom(N,Q)
PxHom(P,Q) -->Q
to the following one
Mx((Hom(M,N)xHom(N,P))xHom(P,Q))--> Mx(Hom(M,N)x(Hom(N,P)xHom(P,Q)))
NxHom(N,P)xHom(P,Q) MxHom(M,N)xHom(N,Q)
PxHom(P,Q) NxHom(N,Q)
Q = Q
(and after that the arrows your wrote before become diagonal like in
the case of homotopy).
This indeed resembles very much what we need, the diagram with Homs,
but how do you organize the passage from your diagram to the diagram
with Homs? The transformation C(LxM,N)->C(L,Hom(M,N)) is not a
functor, it is a natural transformation of functors, so we cannot
apply it to our diagram directly... There must be a trick, that people
in your science use in such situations...
Sergei
P.S. With your permission, I would prefer to contact you directly, by
e-mail, because all those diagrams become much more intelligible, when
written in TeX.
P.P.S. An I bet, after reconstructing all the details we shall obtain
something like what I obtained before with another "evident result" -
a proof with numerous diagrams on several pages...
sergei-...@rambler.ru
Thank you very much, I understood your proof on receiving the pdf-file
by e-mail. I am sorry for being stupid.
I want to thank also the organizers of this forum - this was indeed a
great idea to help mathematicians to communicate with each other. I am
impressed. Unexpectedly fruitful.
Thanks to everyone,
Sergei