Giorgio Mossa
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 Jul 3 revised A functor preserves a product of $A$ and $B$ iff $F(A \times B) \cong F(A) \times F(B)$? added 37 characters in body Jul 3 revised A functor preserves a product of $A$ and $B$ iff $F(A \times B) \cong F(A) \times F(B)$? added 270 characters in body Jul 3 answered A functor preserves a product of $A$ and $B$ iff $F(A \times B) \cong F(A) \times F(B)$? Jun 29 answered Prime ideals in $R[x]$, $R$ a PID Jun 27 comment Natural Transformation: Direct Products @A.P. now that I'm thinking better I realize I should have said 2-comma category: the category $\text{Fam}(\mathbf C)$ is the comma category $(i \downarrow \hat{\mathbf C})$ where $\hat {\mathbf C}$ is the constant functor (from the terminal category in $\mathbf C$) that select the object $\mathbf C$ in $\mathbf {Cat}$ and $i \colon \mathbf{Set} \to \mathbf {Cat}$ is obvious embedding. The objects are functors from sets in $\mathbf C$, the morphisms a 2-commutative triangles in $\mathbf{Cat}$. Jun 23 comment Natural Transformation: Direct Products @A.P. functor categories aren't needed to describe $\mathbf{Fam}(\mathbf {C})$, instead you can use comma-categories to describe $\mathbf{Fam}(\mathbf C)$. Functor categories are needed to deal with products (i.e. limits) to easily describe the image of a morphism through $\prod$ using the universal property of products, this is needed because cones are object in a functor category. Jun 23 comment Should I be using combinations or permutations? Consider you problem. You have $26000$ indipendent variables each one can assume either the value $0$ or the value $1$. A solution for your problem should associate to every variable either the value $0$ or the value $1$.... Jun 23 comment Natural Transformation: Direct Products Happy to have been helpful. The hardest part would be to provide the description of the arrow part of the functor $\prod$, you can skip at first, there is an easy way to characterize this functor by looking to a different presentation of the category $\mathbf{Fam}(\mathbf C)$ which make use of comma and functor categories so maybe it's a little too soon for them. Jun 23 comment Natural Transformation: Direct Products By the way I believe that there is a typo in the book, the mappings $\sigma_*$ and $\tilde \sigma$ are not well defined: in order to let $(\sigma_{ij} b_i)$ being an element of $\prod_j C_j$ one have to specify for every $j \in J$ an $i \in I$ such that $\sigma_{ij} \colon B_i \to C_j$ is an $R$-module morphims, instead of requiring that for every $i \in I$ there is a $j \in J$... Jun 23 comment Natural Transformation: Direct Products The existance of the mappings $\sigma_{i,j}$ in the theorem statements can be easily rephrased in the existance of a morphism $\langle f,\sigma\rangle$ in the category $\mathbf{Fam}(\mathbf{Mod}_R)$. Jun 23 comment Category of sets and multi-valued functions I see, thank you @EricWofsey. Jun 23 answered Natural Transformation: Direct Products Jun 23 comment Natural Transformation: Direct Products Could you provide the reference where this results are stated? Jun 23 comment Category of sets and multi-valued functions I think I may understood the difference in this category and the category Rel: it seems that this is due to the fact that the morphism are represented by partial functions instead of ordinary ones. In the definition given above it could happen that a partial function could associates nothing to an element (which is different than associating the empty set to it). Though I'm wondering if that's the only difference between the two categories. Jun 23 revised Category of sets and multi-valued functions corrected and equation Jun 23 comment Category of sets and multi-valued functions I see, since some authors use dom for source my mistake was understandable (I hope :) ). Anyway I don't understand were the definition of dom is used for the composition. Jun 23 comment Category of sets and multi-valued functions Anyway this is not a category composition: a composition need that for every pair of composable morphisms $f$ and $g$ we have that $\text{source}(g \circ f)=\text{source}(f)$, your composition could produce composites where the domain of the composite could be a proper subset of the domain of the first morphism. Jun 23 comment Category of sets and multi-valued functions could you provide an example? Jun 23 answered Category of sets and multi-valued functions Jun 23 comment Category of sets and multi-valued functions What's the difference between the category of sets and (binary) relations and the category of sets and multivalued function? They should be the same category, should be?