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Exercise 31 - Chapter 4

Prove Lemma 4.30. (Hint: remember to use the fact that \( \mathcal{V} is skeletal.)

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Lemma 4.30

Serial composition of profunctors is associative: given profunctors \( \Phi : \mathcal{P} \rightarrow \mathcal{Q} \), \( \Psi : \mathcal{Q} → \mathcal{R} \), and \( \Upsilon : \mathcal{R} → \mathcal{S} \), we have $$ (\Phi.\Psi).\Upsilon = \Phi.(\Psi.\Upsilon) $$

Comments

  • 1.

    Let \(\newcommand{\cat}[1]{\mathcal{#1}}p\in\cat{P},\,s\in\cat{S}\). \begin{align}(\Phi\circ\Psi)\circ\Upsilon(p,s)=&\bigvee_{r\in\cat{R}}(\Phi\circ\Psi)(p,r)\otimes\Upsilon(r,s)\\=&\bigvee_{r\in\cat{R}}\left(\bigvee_{q\in\cat{Q}}\Phi(p,q)\otimes\Psi(q,r)\right)\otimes\Upsilon(r,s)\\\cong&\bigvee_{r\in\cat{R},q\in\cat{Q}}(\Phi(p,q)\otimes\Psi(q,r))\otimes\Upsilon(r,s)\\\cong&\bigvee_{r\in\cat{R},q\in\cat{Q}}\Phi(p,q)\otimes(\Psi(q,r)\otimes\Upsilon(r,s))\\\cong&\bigvee_{q\in\cat{Q}}\Phi(p,q)\otimes\left(\bigvee_{r\in\cat{R}}\Psi(q,r)\otimes\Upsilon(r,s)\right)\\=&\bigvee_{q\in\cat{Q}}\Phi(p,q)\otimes(\Psi\circ\Upsilon)(q,s)\\=&(\Phi\circ(\Psi\circ\Upsilon))(p,s)\end{align}.

    Since \(\mathcal{V}\) is skeletal, the isomorphisms in the above calculations are equalities. So the lemma is proved.

    Comment Source:Let \\(\newcommand{\cat}[1]{\mathcal{#1}}p\in\\cat{P},\,s\in\cat{S}\\). \begin{align}(\Phi\circ\Psi)\circ\Upsilon(p,s)=&\bigvee\_{r\in\cat{R}}(\Phi\circ\Psi)(p,r)\otimes\Upsilon(r,s)\\\\=&\bigvee\_{r\in\cat{R}}\left(\bigvee\_{q\in\cat{Q}}\Phi(p,q)\otimes\Psi(q,r)\right)\otimes\Upsilon(r,s)\\\\\cong&\bigvee\_{r\in\cat{R},q\in\cat{Q}}(\Phi(p,q)\otimes\Psi(q,r))\otimes\Upsilon(r,s)\\\\\cong&\bigvee\_{r\in\cat{R},q\in\cat{Q}}\Phi(p,q)\otimes(\Psi(q,r)\otimes\Upsilon(r,s))\\\\\cong&\bigvee\_{q\in\cat{Q}}\Phi(p,q)\otimes\left(\bigvee\_{r\in\cat{R}}\Psi(q,r)\otimes\Upsilon(r,s)\right)\\\\=&\bigvee\_{q\in\cat{Q}}\Phi(p,q)\otimes(\Psi\circ\Upsilon)(q,s)\\\\=&(\Phi\circ(\Psi\circ\Upsilon))(p,s)\end{align}. Since \\(\mathcal{V}\\) is skeletal, the isomorphisms in the above calculations are equalities. So the lemma is proved.
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