6,431 reputation
1726
bio website poisson.phc.unipi.it/~mossa
location Earth
age 26
visits member for 3 years, 6 months
seen 19 hours ago

I'm math student, in particular I'm interested in algebra, geometry, topology and category theory (especially higher dimensional category theory) and its application in mathematics.


Dec
6
comment Example of an associative binary operation, without identities or inverses.
@Bartek Yeah, but at the time I wrote the answer the OP didn't say anything about commutativity.
Nov
24
comment System of generators and surjective homomorphism
@MartinBrandenburg ops... thanks for pointing out, I'm going to edit immediately.
Nov
23
comment Homotopy Groups for Categories
I wouldn't dare to call this constuction $\pi_1$ of a category, mainly because this is structure distinguish an n-tuple of composable arrows in the category and their composite....
Nov
21
comment Why is this not a category?
A little add: if instead one consider the families $mor_{\mathbb P}(O,Q)$ of decreasing monotone functions these data do not form a category.... well not in the natural way.
Jul
17
comment Proof completion: if $Y$ is a closed term in strong nf, then $Yx$ weakly reduces to a strong nf $Z$
@RoyO. After a deep reading of the book you cited above I think I've finally found a solution to your problem, take a look and let me know if it's not clear :)
Jul
16
comment Algebras of the environment monad
@MartinBrandenburg aaaaaah I see now. I'm sorry I've misread the question, anyway since it seems that the OP seems interested I think it would better leave it....
Jul
16
comment Algebras of the environment monad
@JeffRussell yes, because $(-)^E \circ (-)^E={(-)^E}^E$ so by the isomorphism ${(-)^E}^E \cong (-)^{E \times E}$ (a.k.a. the exponential law) we get the monad structure.
Jul
15
comment Algebras of the environment monad
@MartinBrandenburg The OP asked "Is there a more natural way to describe these things?" I presented such monad as the image of the comonoid through the yoneda embedding, isn't it a different way to describe the monad?
Jul
8
comment Proof completion: if $Y$ is a closed term in strong nf, then $Yx$ weakly reduces to a strong nf $Z$
@RoyO. I'm probably a little rusty, but isn't it true that a reduction of a term has complexity less than the starting term?
Jul
4
comment On the category of Sets as an example of an algebraic category
@DanaeKissinger Ok, then I think my answer said something about that too, let me edit a little bit to make it clear.
Jun
20
comment How do definitions work in Martin-Lof type theory?
Anyway if you have more question I suppose it's better to go in chat :)
Jun
20
comment How do definitions work in Martin-Lof type theory?
Definitions are the rules for the type.
Jun
20
comment How do definitions work in Martin-Lof type theory?
@user18921 Definition of inductive type in a type theory works like the definition of any other type: you state a rule for the introduction of the type (one that state that the constant of language that represent the type is indeed a type) and some constructors, eliminators and computational rules. The only difference is that such rules are required to have a certain format.
Jun
12
comment Types, Sets and Categories
@CristianGarcia $C_0$ is the type, so of course they have all the same type :), at least this is the usual type theoretic definition of category.
Jun
12
comment Using types instead for basic proofs
@ChristianGarcia Are you trying to consider membership as a relations in the proposition as type paradigm?
May
12
comment Ultrafilter Lemma and Dimension Theorem
@AsafKaragila Ah, ok thanks. For me the dimension theorem was the one stating the existance of a basis... guess I was wrong :)
May
12
comment Ultrafilter Lemma and Dimension Theorem
I belive there's a mistake: the dimension theorem for arbitrary vector spaces is equivalent to the axioms of choice while, as you stated above, the ultrafilter lemma is not.
May
10
comment Homotopy equivalence between $X/A$ and $X$?
Touche, my bad ... Too work and sleep make Giorgio a dull boy.
May
10
comment Homotopy equivalence between $X/A$ and $X$?
This is a very well known theorem, you can find it every book of algebraic topology, for instance in Hatcher Algebraic Topology.
May
6
comment Showing that every map $f : S^2 \rightarrow S^1$ is homotopic to the trivial map
@user125103 yes your proof is correct.