Giorgio Mossa
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 Feb 15 answered What are morphisms in the category of sets $\mathbf{Set}$? Feb 12 comment If $N$ is an $R/I$-submodule of $M$ can we view $N$ as an $R$-submodule of $M$? @Edgar $N$ is a $R/I$-submodule of the $R/I$-module...? Feb 8 reviewed Approve Suggestions on how to verify this partial differential equation solution Feb 7 reviewed Approve $f(f(\sqrt{2}))=\sqrt{2}$ then f has a fixed point Feb 7 comment On which structures does the free group 'naturally' act? Of course a free group naturally act on the set of strings on the generators: which is the Cayley actions, nonetheless I suppose you weren't interested in such trivial action. Feb 3 comment There are two types of functors; covariant and contravariant. Is it right? @user112018 I guess so :) Feb 3 comment There are two types of functors; covariant and contravariant. Is it right? @user112018 I've added an Edit, let me know if that solves your doubts. Feb 3 revised There are two types of functors; covariant and contravariant. Is it right? added 977 characters in body Feb 3 answered There are two types of functors; covariant and contravariant. Is it right? Feb 1 comment What does “Arrows are more important than objects” really mean? @mare_nnoem I don't think so. Here why: yoneda lemma holds for every category while the reduction of points to special kind of arrows is something that holds for groups and topological (and other special kind of concrete categories). To be more detailed the point-arrow fact holds in those concrete categories (i.e. categories with a faithful $\mathbf{Set}$-valued functor) the underlying-set functor is representable (and that's not always the case). Feb 1 revised What does “Arrows are more important than objects” really mean? Made some corrections Feb 1 answered What does “Arrows are more important than objects” really mean? Jan 31 comment Are there any examples of vector spaces over non-numerical fields? If not, why not? @twirlobite I see that you didn't like the answer below. Then I guess you should try to make clear what do you mean by non-numerical vector space. I have interpretated as vector spaces whose underlying field isn't either $\mathbb R$, or $\mathbb C$, or something similar. Am I got it right? Jan 30 comment A question about groups as categories. @Fernando As I said above I've assumed that I was working in a different foundational theory than ZFC: one in which we have proper classes as set-like object, and classes-of-classes and so on. :) Jan 30 comment A question about groups as categories. Of course, the problem is that when doing category theory I always suppose to work with an infinite hierarchy of universes (Tarski-Grothendieck theory, or a rich enough type theory). Thanks for pointing out the needed correction :) Jan 30 comment A question about groups as categories. Injectivity on objects is garanted by the fact that if $x \ne y \in \mathbf C$ then $1_x \in F(x) \setminus F(y)$ and $1_y \in F(y) \setminus F(x)$. From this follows that morphisms having different sources or targets are sent in different functions by $F$. So at most parallel morphisms can be identified by $F$, that's not the case since if $f,g \colon x \to y$ and $F(f)=F(g)$ then $f=f\circ 1_X=F(f)(1_x)=F(g)(1_x)=g \circ 1_x=g$. This proves also injectivity on the arrow part, hence $F$ is an embedding. Jan 30 comment A question about groups as categories. @Fernando are you sure about the non existance of the embedding? Consider any category $\mathbf C$ to every object $x \in \mathbf C$ you can associate the set $F(x)=\bigsqcup_{c \in \mathbf C}\mathbf C(c,x)$ and for every $f \colon x \to y$ in $\mathbf C$ you can associate the function $F(f) \colon F(x) \to F(y)$ defined as $F(f)(g) = f \circ g$ for every $g \colon c \to x$. These data should form a functor from $\mathbf C$ which is an embedding. Jan 29 comment A question about groups as categories. One more suggestion (hope you don't mind): I would eliminate the Last remark label, since that's more the answer than a remark :) Jan 29 comment A question about groups as categories. +1 great and complete answer, but maybe you put too much stuff. Your answer seems a little hard to read, may I suggest you try cut off some additional details reference like the last two paragraphs and left those as comments? I think it would improve the question. Jan 27 answered A functor sending monads to monads