Re: universal element for the functor

In article
<3209677.1141325316072.JavaMail.jakarta@xxxxxxxxxxxxxx
orum.org>,
Meg Weiss <megweiss@xxxxxxxxx> wrote:
In article

<15793250.1141316667274.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org>,
Meg Weiss <megweiss@xxxxxxxxx> wrote:

2. When you reply, be sure to quote the message
you are replying
to. On mathforum this can be achieved by clicking
on the "Quote
original" button over the reply box.

Okay, good. Now, you need to find a balance between
quoting everything
and quoting nothing. You want to trim your quotes so
as to include
only what is needed for context. Yes, it sounds like
contradictory
advise; but good posting is like good writing: you
need to find a
balance between putting in too little (and putting
nothing is
invariably "too little") and too much.


A universal element for a functor F is an ordered
pair (u,R)
consisting of a set R and an element u \in F(R)
with following
property:

I guess your functor goes from Set to some
category whose objects
are sets and whose arrows are functions on the
underlying sets...

To any set S and any element s \in F(S) there is
exactly one
function h: R -> S with F(f)(u) = s.

What happened to h? What is f? Nonsense yet again.
This should be

F(h)(u) = s.

Yor are right F(h)(u) = s.



Let the functor F be F(S) = S x S, F(f) = f x f.


1) For 2 = {1, 2},

This is TERRIBLE notation. Your "2" is playing two
roles (either that,
or your set theory is not well-founded). Much better
to let 2 = {0,1}.


prove that (1,2) \in 2 x 2 is a universal
element for the functor F.

This is nonsense as written. According to the
definition, a "universal
element" is an ordered pair, (u,R), where u is an
element of F(R). If
R=2, and u = 1, then you need 1 to be an element of
F(R)= R x R = 2 x 2,
but "1" is not an element of F(R). So this is
nonsense.

Did you notice that?


Are you asking us to do your homework for you? To
explain the
question? To give a hint? What?

This is not a homework.

Sure seemed like you were assigning homework to this
newsgroup. Perhaps next time you will bother to say
things like "I
would like some help" or "Could someone help me with"

If it is not homework, then what is it? Please tell
me you made up the
problems; if you copied them from somewhere, you need
to sue the
author. The statements are a mess of confusing
notation and nonsense
statements.



I got this problem from the book "ALGEBRA"
by S. MacLANE and G. BIRKHOFF.
This book suppose to be easy,
but it heavily uses an universal element.
I study by myself.





What have you managed? What are you confused
about?

I wrote down F(2) = {(1,1), (1,2), (2,1), (2,2)}
and try to define f
as f(1)= 1, f(2)=2 then I have problem.


Since what you are trying to prove is nonsense, it is
hardly
surprising that you are having a problem. Not to
mention the terrible
use of notation, which is confusing

So, let us rephrase and see if we can figure out WHAT
it is you want
to prove, shall we?

Let 2 = {0,1}, and let F:Set -> Set be the functor
that sends each set
R to R x R, and sends each function f:R->T to the
function
fxf:(RxR)->(TxT).

You want to show that ( (0,1),2) is a universal
element for F. Note
that this makes pirma facie sense, since (0,1) is
indeed an element of
F(2)=2x2 = {(0,0), (0,1), (1,0), (1,1)}.

What does that mean? It means that you want to show
that:

For any set S, and every element s in F(S)=S x S,
there is exactly one
function h: 2 -> S such that F(h)((0,1)) = s.

So, let S be any set, and let s = (a,b) be an element
of S x S. You
want to find a function h:{0,1} -> S, such that

F(h)(0,1) = (a,b).

Now, by definition, F(h)(x,y) = (h x h)(x,y) =
(h(x),h(y)).

So, what should h be?

Your example is a bit of nonsense again, because you
seem to be trying
to map to 2.


It seems like what I did was totally out of line.
Yor are right, we have to show ((0,1), 2) is an
universal element.
If (a,b) = (0,0) then h(0) = 0, h(1) = 0.
If (a,b) = (0,1) then h(0) = 0, h(1) = 1.
If (a,b) = (1,0) then h(0) = 1, h(1) = 0.
If (a,b) = (1,1) then h(0) = 1, h(1) = 1.


I don't get (1,2) where it should be.
How should I define f?

You should first review the definitions and make sure
they make
sense. Then you should fix your notation so it is not
such a
mess. Then you might have a chance at solving the
problem.

2) For 3 = {1, 2, 3},

Yet again confusing notation. Rather, set 3 = {0,1,2}
so that "3" does
not play two roles.

prove that (1,2) \in 3 x 3 is not a universal
element for the
functor F

Nonsense again. Universal objects are supposed to be
ordered pairs,
(u,R), where the first entry is an element of F(R).
So in your case,
your universal object should be a pair, where the
first element is an
ordered pair, and the second element is a set that
contains the two
entries of the first.

So in this case, you mean perhaps:


2) For 3 = {0,1,2}, show that ( (0,1),3) is NOT a
a universal element
for F.

So: you want to show that there is a set S and an
element s in SxS for
which either there is NO function h:3->S such that
F(h)(0,1)=s, or
else that there is more than one function h:3->S such
that
F(h)(0,1)=s. Doing the first problem should give you
a clue that the
problem is that there may be more than one h, since
the condition you
have, namely, F(h)(0,1)=s, will only determine the
value of h on 0 and
on 1, leaving you free to pick the value of 2. So
just make sure your
set S is such that it allows you to define two
different function
h:{0,1,2}->S which have the same value on 0 and 1;
that should tell
you how to go about it.


For example h(0)= 1 h(1) = 2, h(2) = 0,
and h'(0) = 1, h'(1) = 2, h'(2) = 1.

(1,1) can be written (h(0),h(0)) or (h'(0), h'(2))


3) For 1 = {1}, prove that (1,1) \in 1 x 1 is not
universal for F.

More nonsese. WHERE are you getting these statements
from?

Probably rephrasement:

3) For 1 = {0}, prove that ( (0,0), 1) is not
not universal for F.

In this case the problem is going to be that there
are sets S and
elements s in S x S for which there is NO function
h:1->S for which
F(h)(0,0)=s. It should be easy to see why, once you
get the hang of
the first two problems.

Why is this so?
Doesn't h(0) = 0 give desired function?






--
======================================================
================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================
================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

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