* Re: Question on (co)monoids
2013-03-12 13:58 Question on (co)monoids claudio pisani
@ 2013-03-15 14:22 ` Jeff Egger
[not found] ` <E1UFmQ1-0000TM-Ay@mlist.mta.ca>
1 sibling, 0 replies; 3+ messages in thread
From: Jeff Egger @ 2013-03-15 14:22 UTC (permalink / raw)
To: claudio pisani, categories
Hi Claudio,
I was thinking yesterday night about the problem you posed,
> If C is a symmetrical monoidal category and every object has a natural monoid
> structure (that is any map is a monoid morphism) then C is cocartesian monoidal
> (tensor = sums).
in light of the three very different answers which arrived to it on
Wednesday. I was sufficiently surprised by my own conclusions that
I feel a need to share them.
The statement, as given, is a little vague; I interpret it as follows.
PROPOSITION 0: Let (V,@,I) be a symmetric monoidal category, and
suppose that the forgetful functor U : Mon(V,@,I)--->V is split epi in
the (1-)category of (mere) categories and functors. Then I is initial
and @ is cocartesian.
[A splitting of U maps each object of V to a specific monoid,
A |-> (A,m_A,e_A), and every arrow to itself. In other words, every
arrow f:A-->B must be a homomorphism with respect to the specific
structures (A,m_A,e_A) and (B,m_B,e_B).
It turns out, to my surprise, that Proposition 0 is false: I will
momentarily demonstrate this by means of a concrete counter-example;
but first let me address the issue of what is true.
Chris Heunen suggests patching the statement of Proposition 0 as follows.
PROPOSITION 1: Let (V,@,I) be a symmetric monoidal category, and
suppose that the forgetful functor U : Mon(V,@,I)--->V is invertible
in the (1-)category of (mere) categories and functors. Then I is
initial and @ is cocartesian.
Proposition 1 is indeed true, but its utility is somewhat suspect.
To show that U is invertible one must show not only that every object
A carries a monoid structure, but that every monoid structure on A
equals the given one.
Mike Shulman suggests instead (what amounts to):
PROPOSITION 2: Let (V,@,I) be a symmetric monoidal category, and
suppose that the forgetful functor U : Mon(V,@,I)--->V is split epi in
the (1-)category of tensor categories and functors. Then I is initial
and @ is cocartesian.
Not only is Proposition 2 true, but its truth implies that of
Proposition 1: the hypothesis of Proposition 2 requires that m_I and
e_I be the canonical isos, and that m_{A@B} and e_{A@B} be related to
m_A@m_B and e_A@e_B in the usual way; namely, via the canonical isos
A@A@B@B-->A@B@A@B and I-->I@I, respectively. This evidently follows
from the hypothesis of Proposition 1.
But it turns out that the hypothesis of Proposition 2 is still much
stronger than is required. For instance, those parts which refer to
m_I, e_I, and e_{A@B} are entirely superfluous, and the part referring
to m_{A@B} is only used to establish the following property, which
does not even refer to the symmetry of (V,@,I).
(*) id_A@e_B@e_A@id_B splits m_{A@B}
(modulo the canonical iso A@B-->A@I@I@B)
Hence we arrive at the following.
PROPOSITION 3: Let (V,@,I) be a monoidal category, and suppose that
the forgetful functor U : Mon(V,@,I)--->V is split epi in the
(1-)category of (mere) categories and functors. If (*) holds, then I
is initial and @ is cocartesian.
Now I will not write out a proof of Proposition 3, which is a tedious
exercise known (at least in spirit) to many. But I will demonstrate
the necessity of (*) by means of the example promised above.
Let V=Set, I=0, and @ be the tensor product defined by
A@B = A + B + AxB.
This ``unusual'' symmetric monoidal structure on Set was discussed on
the list a few years ago in a thread initiated by Peter Selinger.
I don't remember whether it was mentioned at the time, but monoids in
(Set,@,I) are the same thing as semigroups in (Set,x)---i.e.,
semigroups in the most ordinary sense of the word. For if m : AxA-->A
is associative, then so is [id_A,id_A,m] : A@A-->A. Moreover, the
unique map 0-->A is indeed a unit for any map A@A-->A of the form
[id_A,id_A,f]. Conversely, every monoid in (Set,@,0) is of this form.
In this manner, we obtain an isomorphism between Mon(Set,@,I) and Sgp
in the 1-category of mere categories, which, moreover, commutes with
the two ``underlying set'' functors. But U:Sgp-->Set is a split epi
that is not invertible; for instance, one has the ``left band
functor'' Set-->Sgp which assigns to each set A, the semigroup (A,p_l)
with p_l(a,b)=a. (There is, of course, also a ``right band functor''
which also splits U.)
This is the promised counter-example to Proposition 0. It is not a
counter-example to Proposition 3, however, because the left band
functor violates (*). Let f_{A,B} denote the endomorphism of A@B
defined by composing the following three arrows:
the canonical iso A@B-->A@I@I@B
id_A@e_B@e_A@id_B
m_{A@B}
---then f_{A,B} is a non-trivial idempotent on A@B (not the identity,
as demanded by (*)). Specificially, it maps a pair (a,b) in third
summand of A@B to its first component a in the first summand of A@B.
Note that A+B is the split of this idempotent.
(Obviously, the right band functor also violates (*).)
This situation is typical: in fact, it is easy to show that the maps
f_{A,B}, as defined above, are always idempotents; moreover, the
following generalisation of Proposition 3 also holds.
PROPOSITION 4: Let (V,@,I) be a monoidal category, and suppose that
the forgetful functor U : Mon(V,@,I)--->V is split epi in the
(1-)category of (mere) categories and functors. Then I is initial,
and if each of the idempotents f_{A,B} is split by some object
S_{A,B}, then V has coproducts given by S_{A,B}.
In general, I guess a monoidal category (V,@,I) for which the
forgetful functor U : Mon(V,@,I)--->V is split epi is what a computer
scientist might call a ``model of sum types without beta-reduction''?
I.e., there are maps fst : A-->A@B and snd : B-->A@B and a copairing
operation [,] satisfying fst[a,b]=a, snd[a,b]=b, but not generally
[fst c,snd c]=c.
That is all.
Cheers,
Jeff.
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