Re: universal element for the functor

In article <3209677.1141325316072.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
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.

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.



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.


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 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.

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 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.


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

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.