6,021 reputation
1623
bio website poisson.phc.unipi.it/~mossa
location Earth
age 26
visits member for 3 years, 1 month
seen 9 mins 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.


Oct
10
comment Structure of maximal ideals of the quotient $\mathbb{C}[x,y,z]/ I$
@Sandra Sure: for every point $p \in \mathbb A_K^n$ the ideal $I(p)=\langle x_1 - p_1,\dots,x_n - p_n\rangle$, where by $p_i$ I mean the $i$-th coordinate of $P$. This determinate the generator in $K[x_1,\dots,x_n]$ and so you can apply this result in the case $K= \mathbb C$. Then passes the generators to quotient.
Oct
10
answered Retraction and deformation of P2
Oct
10
answered Structure of maximal ideals of the quotient $\mathbb{C}[x,y,z]/ I$
Oct
10
comment Understanding cohomology with compact support
@Craig here's Hatcher book, I very good book to start learning some algebraic topology math.cornell.edu/~hatcher/AT/ATpage.html on page 251 (if I'm not mistaken) you can find a part about co-homology with compact support.
Oct
10
comment Understanding cohomology with compact support
No it doesn't depend on the choice of the compact, it must just exists a compact subset of $X$ out which vanishes in the sense said above. :) Anyway I've edited the answer again.
Oct
10
revised Understanding cohomology with compact support
made some correction
Oct
10
revised Understanding cohomology with compact support
improved answer
Oct
10
comment Understanding cohomology with compact support
Yes, that's correct: I've just noted that there was a little mistake anywayI'm gonna add some details. So stay tuned.
Oct
10
revised Understanding cohomology with compact support
made a correction
Oct
10
comment Understanding cohomology with compact support
Yes is correct. Just one remark clearly the requirement is that it must be a compact $K$ such that the cochain vanishes outside that compact, and this compact can be different for any compactly supported cochain.
Oct
10
answered Understanding cohomology with compact support
Oct
10
answered Members of equivalence classes in $Z(p^\infty)$
Oct
8
comment What is a good way to think of Factor Groups?
@David I have added the details of the proof that the order of $2+\langle 12 \rangle$ is $6$.
Oct
8
comment What is a good way to think of Factor Groups?
apparently hunger can help in making mistakes :P, it should be correct now.
Oct
8
revised What is a good way to think of Factor Groups?
Made correction
Oct
8
comment What is a good way to think of Factor Groups?
@TobiasKildetoft yes sorry, thanks for pointing out.
Oct
8
answered What is a good way to think of Factor Groups?
Oct
8
comment Homology of wedge sum is the direct sum of homologies
@Fanni you're welcome
Oct
8
answered Homology of wedge sum is the direct sum of homologies
Oct
8
comment Why does Munkres define functions in a seemingly complicated way?
@GrumpyParsnip Ah, sorry I didn't get that, my bad :P