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# David Lambert

• Now for the second snake equation (following the above calculation and making the necessary changes): \begin{align}\rho_{\cat{X}\op}\circ(U_{\cat{X}\op}\times\epsilon_\cat{X})\circ\alpha_{\cat{X\op,X,X\op}}\circ(\eta_\cat{X}\times U_{\cat{X}\op})\ci…
• Note that since $$\newcommand{\op}[]{^\mathrm{op}}\cat{X}\op\neq\cat{X}$$, both snake equations must be checked. The first: \begin{align}\lambda_\cat{X}\circ(\epsilon_\cat{X}\times U_\cat{X})\circ\alpha_{\cat{X,X\op,X}}^{-1}\circ(U_\cat{X}\times\et…
• In order to check the snake equations, one must first define an associator profunctor. As in the previous exercise, this will be defined in terms of a functor $$\newcommand{\Ob}{\mathrm{Ob}(#1)}\newcommand{\Op}{(#1)^\mathrm{op}}\newcommand{\c… • Working on the next problem, I realized that my solution above is wrong. In particular, \(\lambda_\cat{X}\circ\lambda^{-1}_\cat{X}=U_{\cat{X}\times\textbf{1}}$$ and $$\lambda^{-1}_\cat{X}\circ\lambda_\cat{X}=U_\cat{X}$$ should hold (by definition o…
• $$\newcommand{\cat}{\mathcal{#1}}\def\op{{^\mathrm{op}}}\newcommand{\Ob}{\mathrm{Ob}(#1)}\cat{X}$$ and 1 are $$\cat{V}$$-categories. A morphism in the profunctor category over $$\cat{V}$$, i.e., $$\textbf{Prof}_\cat{V}$$, is a $$\cat{V}$$-pro…
• $$\require{AMScd}$$ \begin{CD} \text{EM}_6\unicode{10227}\tt{Bob} @>\text{EM}_1>> \tt{Grace} \\@V\text{EM}_3VV {}@VV\text{EM}_2V \\ \tt{Emory} @. \tt{Pat} \end{CD} \begin{array}{ccc} \tt{Sue}&\unicode{8644}_{\text{EM}_5}^{\text{EM}_4}&a…
• \[ \require{enclose} \def\uline#1#2{\enclose{updiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} \def\dline#1#2{\enclose{downdiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} % \def\place#1#2#3{\smash{\rlap{\hskip{#1em}\raise{#2em}{#3}}}} \begin{…
• I found a discussion on Stack Exchange that allows one to draw figures like that below. However, the process seems to require some fiddling in general to make everything look nice.\[ \require{enclose} \def\uline#1#2{\enclose{updiagonalstrike}{\phan…
• In a Lawvere metric space, $$I=0$$. So, an identity element is a map $$\mathrm{id}_x:0\to\mathcal{X}(x,x)$$. In terms of distances, this means that $$0\geq d(x,x)$$ (since $$0$$ is the smallest element of $$[0,\infty]$$, this means $$d(x,x)=0$$ (t…
• Requirement 1. here corresponds to requirement 1. in the definition of a category. The hom-object becomes the set of morphisms. $$\mathrm{id}_x$$ picks out a distinguished element of the set of morphisms from an object to itself, which by the defi…
• $$g_E(5,3)=\mathrm{false}$$, $$g_F(5,3)=2$$. $$g_E(3,5)=\mathrm{true}$$, $$f_F(3,5)=-2$$. $$h(5,\mathrm{true})=5$$. $$h(-5,\mathrm{true})=-5$$. $$h(-5,\mathrm{false})=6$$.  $$q_G(-2,3)=2$$, $$q_F(-2,3)=-13$$. $$q_G(2,3)=-1$$, $$q_F(2,3)=7$$…
• $$\newcommand{\cat}{\mathcal{#1}}\def\Ob{{\mathrm{Ob}}}\Ob(P)$$ serves as the set on which the preorder is defined. The functor $$\otimes:\cat{P}\times\cat{P}\to\cat{P}$$ induces a function $$\otimes_\Ob:\Ob(\cat{P})\times\Ob(\cat{P})\to\Ob(\cat… • Suppose \(\def\cat#1{{\mathcal{#1}}}$$ $$\def\comp#1{{\widehat{#1}}}$$ $$\def\conj#1{{\check{#1}}}$$ $$\def\id{{\mathrm{id}}}\comp{F}=\conj{G}$$. By the definitions of companion and conjoint, $$\cat{Q}(F(p),q)=\comp{F}(p,q)=\conj{G}(p,q)=\cat{P}(p… • Its conjoint is the function that sends \((d,a,b,c)$$ to $$\mathrm{true}$$ if $$d\leq a+b+c$$ and to $$\mathrm{false}$$ otherwise.
• By definition of the identity $$\newcommand{\idprof}{\mathrm{id}(#1)}\newcommand{\cat}{\mathcal{#1}}\newcommand{\companion}{\widehat{#1}}\cat{V}-$$functor, $$\cat{P}(\idprof{p},\idprof{q})=\cat{P}(p,q)$$. By the definitions of companion (,…
• Let $$\newcommand{\cat}{\mathcal{#1}}p\in\cat{P},\,s\in\cat{S}$$. \begin{align}(\Phi\circ\Psi)\circ\Upsilon(p,s)=&\bigvee_{r\in\cat{R}}(\Phi\circ\Psi)(p,r)\otimes\Upsilon(r,s)\\=&\bigvee_{r\in\cat{R}}\left(\bigvee_{q\in\cat{Q}}\Phi(p,q)\…
• In the current version of Seven Sketches (in Compositionality), there is a third part: 3 Justify each of the three steps $$(=,\leq,\leq)\newcommand{\cat}{\mathcal{#1}}$$ in \eqref{SecondHalfLeftUnitLaw}. \begin{equation}\label{SecondHalfLeftUnit…
• $$\Phi(p,q)=I\otimes\Phi(p,q)$$ holds because $$\mathcal{V}$$ is a symmetric monoidal preorder (in particular, this is property b) in the definition of a symmetric monoidal structure on a preorder). $$I\otimes\Phi(p,q)\leq\mathcal{P}(p,p)\otimes\P… • From comment #28: So after working that out, I think the inverse functor is just the dual functor for a linear map. If we have two different vector spaces \(x \text{ and } y \text{ where } Ax =y$$ Since A is invertible, $$A^{-1}A = I$$. S…
• Let $$x\in\uparrow p$$, $$y\in P$$, and $$x\leq y$$. By definition of $$\uparrow p$$, $$p\leq x$$. By transitivity, this implies that $$p\leq y$$. So, $$y\in\uparrow p$$. $$P^\mathrm{op}$$ has the same elements as $$P$$, so $$\uparrow$$ maps $$P… • \(\require{begingroup}\begingroup\require{AMScd}\newcommand{\tlap}{\raisebox{0pt}[0pt][0pt]{#1}}\newcommand{\blap}{\vbox to 0pt{\hbox{#1}\vss}}$$ If you only need horizontal or vertical arrows or equalities, you can use AMScd. To load it put …
• Ah, ok, I see now that I should have used the function on objects induced by the profunctor $$\cap_\mathcal{X}$$. So, the first diagram chase should say: Let $$x\in\mathcal{X}$$. [ \lambda_\mathcal{X}^{-1}(x)=1\otimes x.] [ ( \cap_\mathcal{X}\ot…
• Let $$x\in\mathcal{X}$$. [ \lambda_\mathcal{X}^{-1}(x)=I\otimes x.] [ ( \cap_\mathcal{X}\otimes1_\mathcal{X})(I\otimes x)=\bigvee_{y\in\mathcal{X}}(y\otimes y^\mathrm{op})\otimes x .] [ \alpha_\mathcal{X,X^\mathrm{op},X}(\bigvee_{y\in\mathcal{X}…
• $$(\check\Sigma\circ (\mathrm{Sell} \times id_\)\circ (\mathrm{Purchase}\times id_\)\circ\hat\Sigma)(\nu,\mu)=\check\Sigma\circ(\mathrm{Sell}\times id_\\circ\mathrm{Purchase}\times id_\)\circ\hat\Sigma(\nu,\mu)$$ leads to $$\hat\Sigma(\alpha,\be… • I would like to solve Puzzle 223 by introducing notation I am more familiar with, as a physicist. I will represent the feasability relations in Einstein notation. I will represent \(\mathrm{True}$$ and $$\mathrm{False}$$ by 1 and 0, respectively. …
• I like your approach, and I think your proof is more elegant than mine (and definitely better written).
• $$F^\mathrm{op}$$ should do the same thing as $$F$$ on objects (i.e., $$\forall c\in\mathcal{C}\,\,F(c)=F^\mathrm{op}(c)$$). On morphisms, I imagine it should take $$f^\mathrm{op}$$ to $$(F(f))^\mathrm{op}$$. This works trivially on identities. \…
• Via Theorem 3.90: $$V=\{1\}$$. $$\mathrm{lim}_\textbf{1}D=\{(d_1)|d_1\in D(1)\,\mathrm{and}\,D(\mathrm{id}_1)(d_1)=d_1\}$$ together with a projection map $$p_1((d_1))=d_1$$. However, since $$D$$ is a functor, the constraint is always satisfied, a…
• A terminal object in Cone$$(X,Y)$$ is a pair of morphisms $$X\xleftarrow[]{\pi_1}X\times Y\xrightarrow[]{\pi_2}Y$$ such that for every pair of morphisms $$X\xleftarrow[]{f}C\xrightarrow[]{g}Y$$, there is a unique morphism $$(f,g):C\to X\times Y$$ su…
• The products of identity morphism for pairs of objects, one from $$\mathcal {C}$$ and the other from $$\mathcal {D}$$. Because composition in each multiplicand is associative. It has objects $$(1,a)$$, $$(1,b)$$, and a single non-identity morphism:…