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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
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
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
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?
Jun
23
comment Proving consistency by constructing models? How and why?
@NikolajK that's an interesting question. In my knowledge no one has build an inconsistent set theory after Russell paradox. History tells us that every kind of inconsistency is usually a variation of Russell paradox and is based on self-referentiality: a statement that asserts that is true or false. Since the foundational crisis mathematicians and logicians have learnt to develop systems that do not allow to derive statements of this kind, in this way they builded systems which hopefully should be consistent.
Jun
22
comment Does mathematics become circular at the bottom? What is at the bottom of mathematics?
@user119615 I've made edits to the answer, hoping it's more clear now.
Apr
24
comment Zorn's Lemma Application for Finding Maximal Submodule?
Since $\bar L$ is a maximal element in $\mathscr{F}$ you have that for every other element $L \in \mathscr{F}$, that is every other submodule of $M$ such that $x \not \in L$ and $N \subseteq L$ you have that if $L \supseteq \bar L$ then $L = \bar L$. This is the definition of maximal element of a poset.
Apr
24
comment Zorn's Lemma Application for Finding Maximal Submodule?
Because you're element belong to $\mathscr{F}$ and every element of this poset has the desidered property.
Mar
26
comment Limit as universal arrow
@LuigiM forgive me, in the first part of the answer I've got confused... I've edited to correct the mistake. :)
Feb
15
comment The functor $\mathbf{D} \rightarrow \mathbf{Prof}$ obtained by “splitting” $F : \mathbf{C} \rightarrow \mathbf{D}$ at each object of $\mathbf{D}.$
@goblin it is indeed, the part I'm referring to is the last section Distributors and generalized fibrations.
Jan
8
comment Degree of the extension $\mathbb{Q}(\zeta_3,\zeta_7)$ over $\mathbb{Q}$
@JackD'Aurizio for start since $[\mathbb Q(\zeta_7):\mathbb Q]=6$ it's not so clear why $\sqrt{-3}=Q_1+Q_2\zeta_7$ and not $Q_0+Q_1\zeta_7+\dots+Q_6 \zeta_7^6$, second it's not so clear (at least not to me) why the relation $-3=(Q_1+Q_2\zeta_7)^2$, which holds in $\mathbf Q(\zeta_7)$ should also hold in the finite field $\mathbb F$.... guess I'm a little rusty on arithmetics...