## Mateship

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### Mateship

- Se incontri il Buddha uccidilo. Devi vivere libero da ogni dogma: se non riesci a uccidere Buddha, come ucciderai il tuo pregiudizio?
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killing_buddha

Messaggi: 2615
Iscritto il: gio 17 lug 2008, 19:51

### Re: Mateship

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"

killing_buddha

Messaggi: 2615
Iscritto il: gio 17 lug 2008, 19:51

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