Limitses. Direct ones.

t.b.

Apr 16, 2012 22:35

lemme debug on main :)

$$\DeclareMathOperator{\Hom}{Hom}\begin{align*}\Hom{\left(\left(\bigoplus M_i \right) \otimes M, N\right)} &=\Hom{\left(\bigoplus M_i, \Hom{(M,N)}\right)} = \prod \Hom{(M_i,\Hom{(M,N)})} \\ & = \prod \Hom{(M_i \otimes M, N))} = \Hom{\left(\bigoplus (M_i \otimes M),N\right)}\end{align*}$$

Benjamin Lim

ok

so

t.b.

and replacing $\bigoplus$ with $\varinjlim$ and $\prod$ with $\varprojlim$ you get the result that tensor products commute with direct limits from Yoneda.

Benjamin Lim

@tb I don't know about how hom relates with the direct product, as well as inverse limits

you see theo because your mathematics is at a much higher level than me you can see all these connections

t.b.

Well, $\Hom{\left(\bigoplus M_i, N\right)} = \prod \Hom{(M_i, N)}$. Let us think about what that really says.

Giving an element of the left hand side, is just giving a morphism $f: \bigoplus M_i \to N$.

Benjamin Lim

ok

yes

ah ok

that

t.b.

Apr 16, 2012 22:42

Now giving an element on the right hand side ...

Benjamin Lim

a morphism like that is just a direct sum of morphisms from $M_i$ to $N$?

t.b.

Well, $\bigoplus M_i$ is a module and you have the inclusions $j_i: M_i \to \bigoplus M_i$.

Benjamin Lim

that just says

yes

do by the universal property of the direct product

t.b.

Now $\left (f \circ j_i\right)_{i \in I} \in \prod \Hom{(M_i,N)}$, right?

Benjamin Lim

yeah

t.b.

Apr 16, 2012 22:45

Conversely (now comes the universal property):

Given $(g_i)_{i \in I} \in \prod \Hom{(M_i,N)}$

Benjamin Lim

wait wait

hold on

t.b.

sure

Benjamin Lim

what is the map that sends an element on the left to one on the right

let me figure that out

t.b.

(I gave that in the two messages before "Conversely" :))

Benjamin Lim

yes

wait let me digest

I am not so fast

t.b.

Apr 16, 2012 22:48

The map is $f \mapsto \left( f \circ j_i \right)_{i \in I}$ where $j_i : M_i \to \bigoplus M_i$ are the inclusions.

Benjamin Lim

ok

well

hhhhhh

oh ok

we get the element $(fj_1,fj_2,\ldots fj_n.....)$ in the direct product

possibly all terms non-zero

t.b.

exactly.

Benjamin Lim

right

t.b.

Ready for the other direction?

Benjamin Lim

yes

t.b.

Apr 16, 2012 22:51

So, given an element $(g_i)_{i \in I} \in \prod \Hom{(M_i,N)}$

That, is given $g_i: M_i \to N$ for all $i \in I$.

Benjamin Lim

then $g_i : M_i \rightarrow N$

t.b.

(your turn)

Benjamin Lim

By the universal property of the direct sum

there is precisely one linear map

$g : \bigoplus M \to N$

t.b.

such that...

Benjamin Lim

defined by $g( ( m_i)_{i \in I}) = \sum g_i(m_i)$

whoops

wait

Now the sum makes sense because all but finitely many terms are zero

t.b.

Apr 16, 2012 22:54

maybe better to write $g \left( \sum_{i \in I} m_i\right) = \sum g_i(m_i)$

Benjamin Lim

yeah ok

t.b.

And even better to write: there's a unique $g \in \Hom{\left(\bigoplus M_i, N\right)}$ such that $g_i = g \circ j_i$ for all $i \in I$.

Benjamin Lim

yeah

so you used $j_i$ for the canonical inclusions, ok

okokokokokokokokok

now that is good

@tb So in your chain of equalities before you applied yoneda

t.b.

Now these maps: $\Phi: f \mapsto \left( f \circ j_i\right)_{i \in I}$ and $\Psi: (g_i)_{i \in I} \mapsto g$ such that $g_i = g \circ j_i$ are mutually inverse.

Benjamin Lim

oh crap forgot to check that!!

t.b.

Apr 16, 2012 22:57

That's what the universal property tells you!

Benjamin Lim

oh ok

yes

t.b.

So $f = \Psi(\Phi(f))$ comes from the uniqueness of the map $g$ such that blabla.

Conversely, $(g_i)_{i \in I} = \Phi(\Psi((g_i)_{i \in I}))$ by how $g$ is determined...

Benjamin Lim

hey theo

t.b.

yes, ben?

Benjamin Lim

you see that map $g$ was determined by just one $g_i$ right

wait

t.b.

Apr 16, 2012 23:01

No, there was a for all $i \in I$.

Benjamin Lim

so the point $(g_1, g_2, \ldots....)$ was just collapsed to $g$

sorry I am not assuming $I$ is countable or anything just wanna show

that the sequence of $g's$ is collapsed to $g$

t.b.

I see what you mean.

After all, we want to prove an equality $\Hom(\ldots) = \prod \Hom(\ldots)$

Benjamin Lim

yes

t.b.

on the left hand side you have just one $\Hom$ thing, and on the right hand side a product of $\Hom$-things. So that's how it should be :)

N3buchadnezzar

Heya, mind anyone help me find a post in the site?

Benjamin Lim

Apr 16, 2012 23:03

yes

N3buchadnezzar

I am looking for a closed form of a integral

Benjamin Lim

that's what I was going to note

Jonas Teuwen

@N3buchadnezzar Again?

Benjamin Lim

One thing on the left was sent to a product of things on the right

okook

N3buchadnezzar

$$ \int_0^\infty \frac{x^{n-1}}{1+x^m} \, \mathrm{d}x$$

I know it is lurking around on the site, but I can not remember where. Nor how to search for it. I know anon posted an answer ..

ops

Benjamin Lim

Apr 16, 2012 23:06

@tb Right

Here's my go at why $f = \Psi(\Phi(f))$

Jonas Teuwen

@N3buchadnezzar That's just a hypergeometric function.

t.b.

I'm here.

Benjamin Lim

By the universal property of the direct sum we have a map $g: \bigoplus M_i to N$ such that $g \circ j_i = f \circ j_i$

By injectivity of the $j_i$ this means that $g = f$

N3buchadnezzar

@JonasTeuwen Yes, but it has a closed form. Something with 1/sin(m \pi) or something

Jonas Teuwen

That is closed.

t.b.

Apr 16, 2012 23:08

@BenjaminLim No, the universal property tells you that there's a unique $g$ such that blah blah, and $f$ already does that blah blah.

Benjamin Lim

ah crap

yes

there is a unique map $g$ from $\bigoplus M_i$ to $N$

but then we already have $f : \bigoplus M_i \to N$ that makes the diagram commute

so $ f = g$

t.b.

yes: you start with $\Phi(f) = \left( f \circ j_i\right)_{i \in I}$, get a unique $g$ such that $g \circ j_i = f \circ j_i$ for all $i$, so $g = f$.

Benjamin Lim

yes yes yes yes

t.b.

Now in the other direction: you start with $(g_i)_{i \in I} \in \prod_{i \in I} \Hom{(M_i,N)}$.

Benjamin Lim

that gets sent to the unique linear map $g : \bigoplus M_i \longrightarrow N$

t.b.

Apr 16, 2012 23:12

the universal property gives you a unique $g \in \Hom{(\bigoplus_{i \in I} M_i, N)}$

Benjamin Lim

and this $g$ under $\Phi$ gets sent to $(g \circ j_i)_{i\in I}$

t.b.

such that $\left(g \circ j_i\right)_{i \in I} = (g_i)_{i \in I}$.

and $\Phi(g)$ is by definition ^

(yes)

Hi, Ilya!

Benjamin Lim

yeah so that was easy

@Ilya pryvyet

N3buchadnezzar

@Ilya: Hey

Ilya

@tb: good morning

@Ben: privet

@N3: hi

Benjamin Lim

Apr 16, 2012 23:13

@tb Because $g_i$ by definition is $g \circ j_i$

Ilya

decreasing number of letters

t.b.

@BenjaminLim yes.

Benjamin Lim

@tb So

That will not help me in that exercise because at the end you want to use Yoneda right??

You want to say that $\hom(M,-) \cong \hom(M',-) \implies M \cong M'$

N3buchadnezzar

@Ilya Remember anything about posts in the site regarding $ \int_0^\infty x^{n-1}/\left( 1 + x^m\right)\,\mathrm{d}x$ by any chance ? =)'

t.b.

@BenjaminLim actually we're applying $\Hom{({-},N)}$

Ilya

Apr 16, 2012 23:15

@N3 reminds of Chebyshev integrals a bit, but I rarely deal with such functions

Benjamin Lim

ok

@tb You know what

all these things, we never really used that $M_i$ were modules right?

just some "object"

t.b.

No, it's a purely categorical argument.

Benjamin Lim

yeah

like I said man my view of mathematics has changed very rapidly

@tb Look at this guy employing thermonuclear weapons here: math.stackexchange.com/a/132327/5783

Ilya

@N3 $m,n$ are integers? because in this case the anti-derivative can be found

t.b.

@BenjaminLim I saw that. The edit from divisible to torsion free was due to a comment of mine :)

Benjamin Lim

Apr 16, 2012 23:19

oh ok

@tb Proof that $\Bbb{Z}$ is a PID with topology???????????????????????????

t.b.

Next thing to understand: $\Hom{(M \otimes M', N)} = \Hom{(M,\Hom{(M',N)})}$

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