Mateship

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Mateship

Messaggioda killing_buddha il gio 27 giu 2013, 20:57

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- Se incontri il Buddha uccidilo. Devi vivere libero da ogni dogma: se non riesci a uccidere Buddha, come ucciderai il tuo pregiudizio?
- "Peu d'abstraction on éloigne de la géometrie; beaucoup on y ramène"
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killing_buddha
 
Messaggi: 2622
Iscritto il: gio 17 lug 2008, 19:51

Re: Mateship

Messaggioda killing_buddha il mar 24 set 2013, 17:00

Somebody seems to be interested in the proof, so here it is.

Suppose the arrow $a$ exists; the other case goes in the same way. If an arrow like $b$ exists too, composing it with $-*K_1$ in the right diagram we obtain

    \xymatrix@C=2cm{
H'K_1 \ar@{=>}[r]^{H'*\eta_1*K_1}\ar@{=>}[d]_{\eta_2*H'*K_1} & H'K_1J_1K_1\ar@{=>}[d]\ar@{=>}[r]^{H'K_1*\epsilon_1} & H' K_1\ar@{=>}[d]^\beta \\
K_2J_2H'K_1\ar@{=>}[r]_{K_2 * a * K_1 } & K_2 H J_1 K_1 \ar@{=>}[r]_{K_2H*\epsilon_1}& K_2H
}
where the first square commutes by hypothesis, and the second thanks to the naturality of $\epsilon_1$. But now we have the following chain of equalities between diagrams:

    \xymatrix{
\ar[r]^{J_1K_1}\ar@{=}[d] &\ar@{=}[d]\ar[r] &\ar@{=}[d]&&&\\
\ar@{=}[d]\ar@{=}[r]&\ar[r]_{H'K_1}&\ar@{=}[d]&&&\\
\ar[rr]_{K_2H}&&&&&
\ar@{=>}^{\epsilon_1}(5,-3);(5,-7)
\ar@{=>}_b(10,-14);(10,-18)
} $\qquad$ \xymatrix{
\ar[r]^{J_1K_1}\ar@{=}[d] &\ar@{=}[d]\ar[r] &\ar@{=}[d]&&&\\
\ar@{=}[d]\ar@{=}[r]&\ar[r]^{H'K_1}\ar@{=}[d]&\ar@{=}[d]&&&\\
\ar@{=}[r]&\ar[r]_{K_2H}&&&&&
\ar@{=>}^{\epsilon_1}(5,-3);(5,-7)
\ar@{=>}_b(15,-14);(15,-18)
} $\qquad$ \xymatrix{
\ar[r]^{J_1K_1}\ar@{=}[d] &\ar@{=}[d]\ar[r]^{H'K_1} &\ar@{=}[d]&\\
\ar@{=}[d]\ar[r]^{J_1K_1}&\ar[r]_{K_2H}\ar@{=}[d]&\ar@{=}[d]&&\\
\ar@{=}[r]&\ar[r]_{K_2H}&&&
\ar@{=>}^{\epsilon_1}(5,-13);(5,-17)
\ar@{=>}_b(15,-4);(15,-8)
}
(the third, thanks to the interchange law), or in other words
$$
(H'K_1 * \epsilon_1)\circ (H' * (\eta_1* K_1)) = (H' * (K_1 * \epsilon_1))\circ H' * (\eta_1 * K_1) = H' * ((K_1 * \epsilon_1)\circ(\eta_1 * K_1) ) = 1_{H' K_1}
$$ thanks to the triangles identity applied to $J_1\dashv K_1$. On the other hand, if we now reconsider the first diagram, $ b \circ (\text{upper row})=b=$ the composition
$$
H' K_1 \xrightarrow{\eta_2 * H'K_1} K_2J_2H'K_1\xrightarrow{K_2 * a * K_1} K_2 HJ_1 K_1 \xrightarrow{K_2H * \epsilon_1} K_2 H
$$ So if $b$ exists (and if we define it in this way, it does exists) must be unique.

Now we go for the last paragraph: consider the diagram

    \xymatrix@C=2cm{
J_2 H' K_1 \ar@{=>}[r]^{J_2\eta_2 H'K_1} \ar@{=}[dr]& J_2 K_2 J_2 H' K_1 \ar@{=>}[d]\ar@{=>}[r]^{J_2K_2 a K_1} & J_2 K_2 HJ_1K_1 \ar@{=>}[d]\ar@{=>}[r]^{J_2 K_2 H\epsilon_1} & J_2 K_2 H\ar@{=>}[d]^{\epsilon_2 H}\\
 & J_2 H' K_1 \ar@{=>}[r]_{a K_1} & H J_1 K_1 \ar@{=>}[r]_{H \epsilon_1} & H
}
This is commutative, since

  • The triangle commutes thanks to the zigzag identities of the adjunctions involved:
    $$
    (\epsilon_2 * JH'K_1)\circ (J_2 * \eta_2 * H' K_1) = \big((\epsilon_2 * J_2)(J_2 * \eta_2)\big)\circ H' K_1 = 1_{J_2 H'K_1}
    $$
  • The left square commutes, since we have the following chain of equalities

      \xymatrix{
\ar[r]^{K_1} \ar@{=}[d]&\ar[r]^{H'J_2}\ar@{=}[d] & \ar@{=}[d]\ar[rr]&&\ar@{=}[d]\\
\ar@{=}[d]\ar[r]_{K_1}&\ar[r]_{HJ_1}&\ar@{=}[d]\ar[rr]&&\ar@{=}[d]\\
\ar[rr]&&\ar@{=}[rr]&&
\ar@{=>}^{\epsilon_2} (31,-13);(31,-18)
\ar@{=>}^{a} (15,-3);(15,-8)
} \qquad\qquad 
\xymatrix{
\ar[r]^{K_1}\ar@{=}[d]&\ar@{=}[d]\ar[rr]^{J_2H'}&&\ar@{=}[d]\ar[r]^{K_2}&\ar@{=}[d]\ar[r]^{J_2}&\ar@{=}[d]\\
\ar[r]_{K_1}\ar@{=}[d]&\ar[rr]_{HJ_1}\ar@{=}[d]&&\ar[r]_{K_2}\ar@{=}[d]&\ar[r]_{J_2}&\ar@{=}[d]\\
\ar[r]&\ar[rr]_{HJ_1}& &\ar@{=}[rr]&&
\ar@{=>}^{a} (20,-3);(20,-8)
\ar@{=>}^{\epsilon_2} (42,-13);(42,-18)
}\qquad\qquad
\xymatrix{
\ar[r]^{K_1}\ar@{=}[d]&\ar[rr]^{J_2H' }\ar@{=}[d]&&\ar[rr]^{J_2K_2}\ar@{=}[d]&&\ar@{=}[d]\\
\ar[r]_{K_1}\ar@{=}[d]&\ar[rr]^{J_2H'}\ar@{=}[d]&&\ar@{=}[rr]\ar@{=}[d]&&\ar@{=}[d]\\
\ar[r]&\ar[rr]_{HJ_1}&&\ar@{=}[rr]&&
\ar@{=>}^{a} (20,-13);(20,-18)
\ar@{=>}^{\epsilon_2} (40,-3);(40,-8)
}
    which finally equals $(a * K_1)\circ (\epsilon_2 * J_2H'K_1)$.
  • The right diagram commutes too with a similar argument relative to the composition
      \xymatrix{
H'K_1 J_1 \ar@{=>}[r]\ar@{=>}[d]& K_2 J_2 H' K_1J_1 \ar@{=>}[r]\ar@{=>}[d]& K_2 HJ_1K_1J_1\ar@{=>}[r] \ar@{=>}[d]& K_2 HJ_1\\
H'\ar@{=>}[r] & K_2J_2H'\ar@{=>}[r] & K_2 HJ_1 \ar@{=}[ur]
}
It remains to show that one of the two commutativities implies the other: consider the sequence of pasting diagrams

    \xymatrix{
\ar@{=}[rr]\ar@{=}[d]&&\ar@{=}[d]\ar[r]^{H'}&\ar@{=}[d]\ar[r]^{J_2}&\ar@{=}[d]\\
\ar@{=}[d]\ar[r]_{J_1}&\ar@{=}[d]\ar[r]_{K_1}& \ar[r] &\ar@{=}[d]\ar[r] &\ar@{=}[d]\\
\ar@{=}[d]\ar[r]&\ar[r]_H&\ar[r]_{K_2}\ar@{=}[d]&\ar[r]&\ar@{=}[d]\\
\ar[rr]&&\ar@{=}[rr]&&
\ar@{=>}^{\eta_1} (10,-3);(10,-8)
\ar@{=>}^{b} (20,-13);(20,-18)
\ar@{=>}^{\epsilon_2} (30,-23);(30,-28)
} \qquad\qquad 
\xymatrix{
\ar@{=}[d]\ar@{=}[rr]&&\ar@{=}[d]\ar[r]&\ar@{=}[d]\ar[r]&\ar@{=}[d]\\
\ar@{=}[d]\ar[r]&\ar@{=}[d]\ar[r]&\ar@{=}[d]\ar[r]&\ar[r] &\ar@{=}[d]\\
\ar@{=}[d]\ar[r]&\ar@{=}[d]\ar[r]&\ar[r]&\ar[r]\ar@{=}[d]&\ar@{=}[d]\\
\ar[r]&\ar@{=}[rr]&&\ar[r]&
\ar@{=>}^{\eta_1} (10,-3);(10,-8)
\ar@{=>}^{a} (30,-13);(30,-18)
\ar@{=>}^{\epsilon_1} (20,-23);(20,-28)
}

    \xymatrix{
\ar@{=}[d]\ar@{=}[rr]&&\ar@{=}[d]\ar[r]& \ar[r]&\ar@{=}[d]\\
\ar@{=}[d]\ar@{=}[rr]&&\ar@{=}[d]\ar[r]&\ar@{=}[d]\ar[r]&\ar@{=}[d]\\
\ar@{=}[d]\ar[r]&\ar@{=}[d]\ar[r]&\ar[r]&\ar@{=}[d]\ar[r]&\ar@{=}[d]\\
\ar[r]&\ar@{=}[rr]&&\ar[r]& 
\ar@{=>}^{a} (30,-3);(30,-8)
\ar@{=>}^{\eta_1} (10,-13);(10,-18)
\ar@{=>}^{\epsilon_1} (20,-23);(20,-28)
}\qquad\qquad 
\xymatrix{
&&\\
\ar@{=}[d]\ar[r]& \ar[r]&\ar@{=}[d]\\
\ar[r]&\ar[r]&\\
&&
\ar@{=>}^{a} (10,-13);(10,-18)
}
[...]
- Se incontri il Buddha uccidilo. Devi vivere libero da ogni dogma: se non riesci a uccidere Buddha, come ucciderai il tuo pregiudizio?
- "Peu d'abstraction on éloigne de la géometrie; beaucoup on y ramène"
Avatar utente
killing_buddha
 
Messaggi: 2622
Iscritto il: gio 17 lug 2008, 19:51


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