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}[1]{\mathrm{Ob}(#1)}\newcommand{\Op}[1]{(#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…
\(\newcommand{\cat}[1]{\mathcal{#1}}\def\op{{^\mathrm{op}}}\newcommand{\Ob}[1]{\mathrm{Ob}(#1)}\cat{X}\) 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…
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…
\(\newcommand{\cat}[1]{\mathcal{#1}}\def\Ob{{\mathrm{Ob}}}\Ob(P)\) 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…
By definition of the identity \(\newcommand{\idprof}[1]{\mathrm{id}(#1)}\newcommand{\cat}[1]{\mathcal{#1}}\newcommand{\companion}[1]{\widehat{#1}}\cat{V}-\)functor, \(\cat{P}(\idprof{p},\idprof{q})=\cat{P}(p,q)\). By the definitions of companion (,…
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)\…
In the current version of Seven Sketches (in Compositionality), there is a third part:
3 Justify each of the three steps \((=,\leq,\leq)\newcommand{\cat}[1]{\mathcal{#1}}\) 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…
\(\require{begingroup}\begingroup\require{AMScd}\newcommand{\tlap}[1]{\raisebox{0pt}[0pt][0pt]{#1}}\newcommand{\blap}[1]{\vbox to 0pt{\hbox{#1}\vss}}\)
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…
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. …
\(F^\mathrm{op}\) 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:…