God's in his heaven, all's right with the world.

To my lovely classmate, my best friend forever, and the 1st lady in the SMS99 alumna, Jane (Shadow),

 

I almost stopped update my blog here. But I write this article, specially for you, to tell you that I did not forget my words and I was almost there. kaka.

 

Anything can happen in such a sudden that one can never expect. Ok, when I transfered at the aeroport, the gate right besides my gate was to Rome. I had never expected in my following trip that I would be in a land that Caeser was once in. When I was standing on the Romanesque remains, the miniature of those in Rome, the first thing I could think about is to let you know I have never forgot my words. The fountain, the amphitheatre, the circ, I was, but alone. Wish you can notice this letter and see the pictures I shared. Hepburn was dead, and so was Peck. The hope remains forever, in the past and in the future!

 

Miss you so much. Hope all are going well there.

 

Sincerely Yours,

SmallX

 

A fountain besides the provincial forum.

Caeser stands on his city and overlooks the Mediterranean.

The city Tarracona (not Roma though...)

Roma circ, looks like some construction site =D

Roman amphitheatre.

The city forum.

Ok, this one is not from Roman time. It is Poblet Monastery.

Wanderer.

Do you seek the advanture beyond the treacherous waterfalls?

Do you seek the mythical being that dwells in the unreachable place?

If you do, then you must first find me.

 

你是个冒险家吗，

穿过名叫危险的瀑布，

寻求那深处传说的真相的人吗，

如果是的话，请追寻我。

 

汝は冒険者か？

危険という名の滝を潜り抜け、

その奥に伝説の正体を求めるものか？

ならば我を求めよ。（暗語 in『ふしぎの海のナディア』）

Diagonal Theorem(Cantor's): $X,Y$ be two objects and $f: X\times X\to Y$ be an arrow, s.t. for any arrow $g: X\to Y$ there is an arrow $x: 1\to X$ s.t. $g=f <id;x>$, then for every endomap $\alpha: Y\to Y$ there is an arrow $y: 1\to Y$ s.t. $\alpha y=y$.(Hint:Consider the arrow $\alpha\f\triangle$)

 

Cantor's Argument(1870's):

Since there is an arrow as logical negation for  $2$, the statement about $Set, 2$ in the theorem does not hold.

 

Russell's(1900's) G$\"o$del's(1930's) and $Tarski(1930's) Argument:

Since the statement about $\mathsf{N}, \Omega$ in the theorem holds, $\Omega$ can not be $2$.

(from Conceptual Mathematics)

comment: There are something, and something else.

 

有理想，有追求。我的故事。

 

もりのこえをきこう。とおいほしたちょからふいてくるかぜのきおくを、くものはなし…（月岛しずく in 『耳をすませば』）

 

You'll have to do as you believe. Feeling the mildness of the wind, listening to the sound from the snow, following the heel of the cat, and looking high up the airship drifting through the sky.

Brouwer's, in 1900's.

About topological spaces and continuous maps. Here is another verison.

Lemma. If $A$ is a retract of $X$, and $X$ has fixed point, then so does $A$.

Axiom 1. If $T$ is an object in $\mathsf{C}$, and $a:T\rightarrow A, s:T\rightarrow S$ are two arrows s.t. $ha=js$, then $pa=s$.

Theorem 1. If $\alpha:B\rightarrow A$ satisfies $h\alpha j=j$, then $p\alpha$ is a retraction for $j$. (Hint:Let $T=S$)

Corollary. If $h\alpha=id_B$, then $p\alpha$ is a retraction for $j$. (Hint:Since id_Bj=j)

Axiom 2. If $T$ is an object in $\mathsf{C}$, and $f,g:T\rightarrow B$ are two arrows, then either there is a point $t:1\rightarrow T$ s.t. $ft=gt$ or there is an arrow $\alpha:T\rightarrow A$ s.t. $h\alpha=g$.

Theorem 2. If $f,g:B\rightarrow B$ are two arrows, and $gj=j$, then either there is a point $b:1\rightarrow B$ s.t. $fb=gb$ or there is a retraction for $j:S\rightarrow B$. (Hint:Let $T=B$ and then apply both theorems)

Corollary. If $f:B\rightarrow B$ is an arrows, then either there is a fixed point for $f$ or there is a retraction for $j:S\rightarrow B$. (Hint:Let $g=id_B$)

(from Conceptual Mathematics, by F.W.Lawvere & S.H.Schanuel, Buffalo NY)

comment: Generally, there is always something implicit, which is fixed.

 

无序中蕴含的有序。

 

しんぱいしないで。いちびょうはけっこうです。ゼロではないし、負整数ではない。（赤木律子 in 『新世紀エヴァンゲリオン』）

 

It is generally believed that ``well begun is half done''. There are two implicit meanings of it. One is that once you begin, you will enjoy the working and go on the road to make it. Having a determined target, holding the right tools, and most importantly, striding with the confidence, the only thing left is to implement according to schedule on the wall. Although there do have some incomplete work, which had begun, after carefully examination, one can amazingly find that most of the failures are due to the bad beginning. Then there comes the other meaning, that it is not easy to begin, or at least to begin well. Situations are complicated, targets are vague, even the tools are not so convenient, too much or too few. To figure out each connections in the sophisticated world is beyond the power of human beings, isn't it? Although there are always orders behind those chaotic superficial phenomena. Do we perceive the world, or we create a world? Can we understand the real world, or we can only understand the world we create? What is real, and what is imagination?

 

PS: PVS 4.0 is available. One feature is that it is now open source, WOW. Here is an associated Wiki http://pvs-wiki.csl.sri.com

Theorem. For every object $X$ of $\mathbb{C}$ there is an initial object in comma category $(X\downarrow G)$, $(\eta_X, GF_X)$, then the following holds:

1. this defines a functor $F$ in the opposite direction of $G$.

2. the arrow $\eta_X$ is natural in $X$, $\eta:Id\Rightarrow GF$.

3. For every object $Y$ of $\mathbb{D}$ there is an initial object in comma category $(F\downarrow Y)$, $(FGY,\epsilon_Y)$.

4. the arrow $\epsilon_Y$ is natural in $Y$, $\epsilon:FG\Rightarrow Id$.

 

Theorem. An adjunction from a category $\mathbb{C}$ to a category $\mathbb{D}$ is given by a pair of opposite functors $(F,G)$, s.t. for all $X$ in $\mathbb{C}$ and $Y$ in $\mathbb{D}$, a bijection $\varphi_{X,Y}:\mathbb{D}(FX,Y)\cong\mathbb{C}(X,GY)$ which is natural in both $X$ and $Y$. Then, the following is determined:

1. a natual transformation $\eta\triangleq\varphi(id_F):Id_\mathbb{C}\Rightarrow GF$, which is called the unit.

2. a natual transformation $\epsilon\triangleq\varphi^{-1}(id_G):FG\Rightarrow Id_\mathbb{D}$, which is called the counit.

In addition, each arrow of the natual transformation is universal and the triangular identities hold, i.e. $U\epsilon\circ\eta_U=id_U$ and $\epsilon_F\circ F\eta=id_F$.

 

Theorem. A monad on a category $\mathbb{C}$ is given by $T=\langle T,\eta,\nu\rangle$, where $T:\mathbb{C}\rightarrow\mathbb{C}$ is an endofunctor, $\eta:Id_\mathbb{C}\Rightarrow T$ is a unit, $\nu:T^2\Rightarrow T$ is a multiplication, s.t. the following diagrams commute.

\xymatrix{ T^3\ar@{=>}[r]^{T\mu}\ar@{=>}[d]_{\mu_T} & T^2\ar@{=>}[d]^{\mu} & T\ar@{=>}[r]^{\eta_T}\ar@{=>}[rd]_{id_T} & T^2\ar@{=>}[d]^{\mu} & T\ar@{=>}[l]_{T\eta}\ar@{=>}[ld]^{id_T} \\
T^2\ar@{=>}[r]_{\mu} & T & & T}

Then, every adjunction gives rise to a monad and every monad determines adjunctions.

 

For example, Eilenberg-Moore category $\mathbb{C}^T$ has $(X,h:TX\to X)$ as objects and $id_X$ as id arrows, which is final; Kleisli category $\mathbb{C}_T$ has $X$ as objects and $\eta_X:X\to TX$ as id arrows, which is initial.

(from LFCS Lecture Notes;Wikipedia, the free encyclopedia)

comment: to observe compatibility in another way.

 

年年岁岁花相似，岁岁年年人不同。（《代悲白头翁》刘希夷）

 

なんとなくむかしにあったようだが、いつかふるさとにかえってしまたかとはときどきおもってる。

 

First assignment, first TA duty, first presentation, etc. and all are going well, and nothing is new in some sense. Is that true that all human beings are the same whatever their races are. I remember where I was born, where is a cool place, with snow covering the roof of each house. Let's call it H. The BSc and MSc degree drive me far away from home to a metropolis, which is a busy city where everything is in a rush. Let's call it P. It follows that all Chinese are the same to some extent. Now the PhD degree drives me even farther away from home to a little country, which is a quiet city where the goose and squirrels run everywhere. Let's call it W. It follows that all terrestrials are the same to some extent. More precisely, there are more similarities between H and W than between P and W. That's OK. Next destination, Mars.