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# Fredrick Eisele

• I will make a guess here and claim that the dagger functor is actually a pair of functors: [ \dagger : \mathcal{X} \rightarrow \mathcal{X}^{op} \text{ and } \dagger : \mathcal{X}^{op} \rightarrow \mathcal{X} ] Were it not so, the statement [ \dagge…
• @Bruno You filled out the matrix with paths of length 1 (which is correct). What Happens when you multiply it by itself?
• Consider a slightly modified problem where 3 contains an additional free morphism. In this case there are two distinct natural transformations $$\alpha, \beta : F \rightarrow H$$ where $$\alpha_3 = c \circ a, \beta_3 = c \circ b$$. It first s…
• https://forum.azimuthproject.org/discussion/comment/19418/#Comment_19418
• I see, some care must be taken when constructing a graph suitable for construction of a Free category. When parallel paths are present in the graph the are assumed to be not equal unless there is an equation specifically stating otherwise. The onl…
• Puzzle 131. ... Again, it may help to draw a graph. I wanted to see where the overcount in comment 13 came from. 6:6 of length 0 (the identity transformations) 6:6 of length 1 [black] (the transformation arrows on the graph) 5:6 of length 2…
• Puzzle 135. For any category $$\mathcal{C}$$, what are functors $$F : \mathcal{C} \to \mathbf{1}$$ like? cocone, colimit, coproduct and join?
• Puzzle 134. For any category $$\mathcal{C}$$, what's another name for a natural transformation $$\alpha : F \Rightarrow G$$ between functors $$F,G: \mathbf{1} \to \mathcal{C}$$? Yet again there's a simple answer using concepts you've learned here…
• Addendum base on comment 22. The commuting squares of which $$G \overset{\alpha}{\Rightarrow} G$$ is typical do exist. [ \begin{matrix} \textbf{1} && \textbf{3} \\ 1_1 & \overset{G}{\rightarrow } & 2_3 \\ id_1 \downarrow & \Dow…
• Puzzle 132. For any category $$\mathcal{C}$$, what's another name for a functor $$F: \mathbf{1} \to \mathcal{C}$$? There's a simple answer using concepts you've already learned in this course. [ F \in Obs(\mathcal{C} ) ] Puzzle 133. For any…
• Puzzle 130. Let $$\mathbf{3}$$ be the free category on this graph: How many functors are there from $$\mathbf{1}$$ to $$\mathbf{3}$$, and how many natural transformations are there between all these functors? In comment 11 the candidates for n…
• [ \begin{array}{c | c c c} \Phi . \Psi & p & q & r & s \\ \hline A & ? & 24 = 3+3+5+9+0+1+1+2 & ? & ? \\ B & ? & ? & ? & ? \\ C & ? & ? & ? & ? \\ D & ? & ? & 9 = 9+0 & …
• We can either do the multiplication or we can examine the graph and find the least expensive path between the specified nodes. [ \begin{matrix} \begin{array}{c | c c c c} M_X^2 & A & B & C & D \\ \hline A & 0 & 6 & 3 &a…
• [ ( M * N )(x, z) := \bigvee_{y \in Y} M(x, y) \otimes N(y, z) ] Using (min, +) gives entries of the form... [ ( M * N )(x, z) := \underset{y \in Y}{min} \; M(x, y) + N(y, z) ]
• If we take $$\mathcal{V}$$ to be $$\textbf{Bool} := (\mathbb{B}, \le_{Bool}, \text{true}, \wedge)$$ and $$\mathcal{X}(m, n) := m \le_X n$$, $$\mathcal{Y}(m, n) := m \le_Y n$$, $$\Phi(m, n) := m \le_{\Phi} n$$ , then [ \mathcal{X}(x' , x)…
• Here is a puzzle to think about. How to express all of the SQL semantics categorically? Ask @RyanWisnesky.
• [ \tag{V-profunctor} \Phi : \mathcal{X}^{op} \times \mathcal{Y} \rightarrow \mathcal{V} ] [ \tag{Bool-profunctor} \Phi : \mathcal{X}^{op} \times \mathcal{Y} \rightarrow \textbf{Bool} ] [ \tag{feasibilty relation} \Phi : \mathcal{X}^{op} \times \math…
• Here I show that it is true via a truth table. [ \begin{array}{c c c | c c | c c | c } b & c & d & b \wedge c & ( b \wedge c ) \le d & c \Rightarrow d & b \le ( c \Rightarrow d ) & ( b \wedge c ) \le d = b \le ( c \Righ…
• 2) Write down a profunctor $$\Lambda : \mathcal{X} \nrightarrow \mathcal{Y}$$ and, reading $$\Lambda(x, y) = \text{true}$$ as “my uncle can explain $$x$$ given $$y$$”, give an interpretation of the fact that the preimage of $$\text{true… • 1) Draw the Hasse diagram for the preorder \( \mathcal{X}^{op} \times \mathcal{Y}$$.
• Inspired by matrix rig comment. E := employee D := department S := string m := manager a := secretary [assistant] d := department name f := first name w := works in [ A^0 = \left( \begin{array}{c | c c c} s \rightarrow t & E & D…
• Do you have a git repository? Yes. https://github.com/CategoricalData/fql
• Thanks @MatthewDoty all is now clear.
• In comment there are some ideas. I would like to know the names of these ideas. We have a monoidal skeleton-category labeled $$\mathbf{Matrix}_{skel}$$ and two categories $$\mathbf{Matrix}_{\mathbb{N}}$$ and $$\mathbf{Matrix}_{multiset}$$ …
• $A^4 = \left( \begin{array}{c | cc} s \rightarrow t & x & y \\ \hline x & \lbrace f;f;f;f, \; g;h;f;f, \; f;g;h;f \; f;f;g;h, \; f;h;g;h \rbrace & \lbrace f;f;f;g, \; g;h;f;g, \; f;g;h;g \rbrace \\ y & \lbrace h;f;f;f \;… • \[ A^3 = \left( \begin{array}{cc} \lbrace f;f;f, \; g;h;f, \; f;g;h \rbrace & \lbrace f;f;g, \; g;h;g \rbrace \\ \lbrace h;f;f, \; h;g;h \rbrace & \lbrace h;f;g \rbrace \end{array} \right)$
• Composition and union are right. $A^1 = \left( \begin{array}{cc} \lbrace f \rbrace & \lbrace g \rbrace \\ \lbrace h \rbrace & \emptyset \end{array} \right)$ \[ A^2 = \left( \begin{array}{cc} \lbrace f;f, \; g;h \rbrace & \lbrace f…
• To make the adjacency matrix notion painfully obvious. [ \left( \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right) ] [ \left( \begin{array}{c | cc} s \rightarrow t & x & y \\ \hline x & \lbrace f \rbrace & \lbrace g \rbrac…
• Then I am lost on what to do with (1) and (2). Do you have any insight?
• The latest version of the text had a major relabeling of all labeled items. I believe this was due to the inclusion of Equations in the labeling scheme. So, it is a good thing but it will require renumbering everything. I updated the labels for Ex…