Reputation
6,257
Next privilege 10,000 Rep.
Access moderator tools
Badges
2 10 31
Newest
 Altruist
Impact
~54k people reached

Apr
17
revised Values of the Herbrand quotient
fixed latex, spelling of “Herbrand”
Apr
17
comment Let $X$ be a normal space then there exists a continuous map $f : X → [0, 1]$ such that $f^{−1} (0) = A$
@5xum Is it? The property doesn’t say that the closed sets in $X$ are exactly countable intersections of open sets, so there might be countable intersections of open sets (such as trivial ones) that aren’t closed, right?
Apr
11
comment I have a question about the multiplicative inverse in any field.
It should. If you tell us the field axioms you are using, I’m sure we can point out which one assures $0 ≠ 1$. Anyway, the only ring $R$ with one in which $0 = 1$ is $R = 0$, since $∀x ∈ R\colon x = x·1 = x·0 = 0$.
Apr
11
comment I have a question about the multiplicative inverse in any field.
If $α$ was a multiplicative inverse of $0$, then $α·0 = 1$, but also $α·0 = 0$ (which is true for all elements $a$ of the field since $a·0 = a·(0+0) = a·0 + a·0$ from which you can additively cancel $a·0$ (by adding $-a·0$ to both sides)). Hence $0 = 1$ which is not allowed by the very definition of a field.
Apr
11
answered about hyperplanes in finite fields
Apr
11
comment about hyperplanes in finite fields
Hyperplanes are vector spaces and if $U ⊂ \mathbf F_q^k$ is a ($k-2$-dimensional) proper subspace and $x \notin U$, then $U ∪ \{x\}$ just won’t be a vector space (think of all the sums $x + u$ with $u ∈ U$ you are missing).
Apr
11
comment Definition of split of exact sequence
@ringwith1 You are thinking of this: If $M$ is an abelian group, then for subgroups $A ⊂ M$ and $B ⊂ M$ it is equivalent that $M$ splits as an internal direct sum $M = A \oplus B$ if and only if for every $m ∈ M$ there are unique $a ∈ A$, $b ∈ B$ such that $m = a + b$. In your case, the latter part of the equivalence may be satisfied in some sense, but $ℤ/nℤ$ $(=B)$ just isn’t a subgroup of $ℤ$ $(=M)$ (and even not so if you regard $ℤ/nℤ$ as the subset $\{0,…,n-1\}$ of $ℤ$, because then the addition on $ℤ/nℤ$ is not inherited from the addition in $ℤ$).
Apr
11
comment Definition of split of exact sequence
Why would you think that $ℤ \cong nℤ \oplus ℤ/nℤ$?
Apr
11
comment When is $G\cong\operatorname{End}(G)$?
@Gaussler Yeah, but it doesn’t work for rings without one (like $2ℤ$) either.
Apr
11
comment When is $G\cong\operatorname{End}(G)$?
@Gaussier For $R = ℤ$, $f = \mathrm{id}_ℤ + \mathrm{id}_ℤ$ is not multiplicative (nor does it preserve $1$), because $f(1·1) = 2 ≠ 4 = f(1)·f(1)$ (that’s right – $\mathrm{Ring}$ is not an additive category).
Apr
9
comment Find an automorphism
Find instead two surjective homomorphisms $ℤ → G$ and use the first isomorphism theorem. Alternatively, directly find two different isomorphisms $ℤ/nℤ → G$ which send $1$ to …?
Apr
8
comment Why are functors exact if they preserve all short exact sequences?
@lenticcatachresis I neither speak nor read Spanish but I might just be able to guess my way through this with a dictionary. With David’s answer, though, this will be unnecessary. Many thanks eitherway!
Apr
8
accepted Why are functors exact if they preserve all short exact sequences?
Apr
8
comment Why are functors exact if they preserve all short exact sequences?
@Berci Thanks, so generally $\operatorname{ker}~if = \operatorname{ker}~f$ whenever $i$ is a monomorphism and $\operatorname{img}~fp = \operatorname{img}~f$ whenever $p$ is an epimorphism – which can be proven by checking the universal properties, right?
Apr
8
comment Why are functors exact if they preserve all short exact sequences?
@Jim I know that, but that is hardly enlightening and not very satisfying unless you prove/understand Mitchell’s embedding theorem first which I haven’t and which seems to me too much of an effort for this problem.
Apr
8
comment Why are functors exact if they preserve all short exact sequences?
Thanks! This seems to work for the category of modules, say, where there is a notion of injectivity and kernels/cokernels can be defined as submodules rather than some limits/colimits. Does this approach generalise to arbitrary abelian categories? (Sorry for being so lazy not to think about it in detail.)
Apr
8
accepted The pushout of an open/closed injective map is open/closed
Apr
8
asked Why are functors exact if they preserve all short exact sequences?
Apr
6
comment An upper bound of a complex number
For real numbers $a$ and $t$, $e^{iat}$ is on the unit circle, hence $|e^{iat}| = 1$?
Apr
6
answered Relative homotopy and composition of maps