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comment Finding the kernel of maps between (polynomial) rings
@user26857 I agree that is a lot like shooting a little bird with a cannon, nonetheless it is a way to take familiarity with the instrument of dimension theory..... at least in my personal opinion.
Jul
27
answered Finding the kernel of maps between (polynomial) rings
Jul
12
comment When the unit of a universal property is an isomorphism
@Berci yes you're right :) thanks for pointing out.
Jul
12
revised When the unit of a universal property is an isomorphism
Made a correction
Jul
11
answered When the unit of a universal property is an isomorphism
Jul
8
comment Can all theorems be deduced directly from the ZFC axioms?
Just a pun: a system that allows to derive everything is inconsistent ... I hope ZFC is not.
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.