39,725 reputation
247119
bio website
location
age
visits member for 3 years, 11 months
seen 58 mins ago

Jan
21
answered The image of a morphism between affine algebraic varieties.
Jan
20
comment Categories with no products/coproducts
Note however that the Yoneda embedding $\mathcal{C} \to [\mathcal{C}^\textrm{op}, \textbf{Set}]$ is not guaranteed to preserve colimits.
Jan
20
comment Categories with no products/coproducts
The question is built on the misconception that all categories are concrete categories of mathematical structures of one kind or another. The fact of the matter is that, by moving to a sufficiently large universe if necessary, any fixed category can be embedded in one that is complete and cocomplete. But this is completely abstract nonsense and is not guaranteed to preserve any of the colimits that previously existed.
Jan
20
comment About stalk of sheaves in an unbalanced category
Taking stalks is a left adjoint functor $\textbf{Sh}(X; \mathcal{C}) \to \mathcal{C}$, so $\textbf{Psh}(X; \mathcal{C})$ doesn't come into play.
Jan
20
comment If $\operatorname{dim} M > \operatorname{dim} N$, is there an injective smooth map $M\to N$?
True, but why does no such smooth map exist? That's not obvious.
Jan
20
answered Sheaf epimorphisms
Jan
20
revised About stalk of sheaves in an unbalanced category
deleted 175 characters in body
Jan
20
comment About stalk of sheaves in an unbalanced category
No, of course not. That's already true for sheaves of abelian groups. (Think, for example, about de Rham cohomology.)
Jan
20
comment Composition of a kernel with a cokernel has a normal image
Yes, that's just pullback pasting and pullback cancellation.
Jan
19
comment Composition of a kernel with a cokernel has a normal image
I don't understand your question. This argument requires much more than just pullback pasting.
Jan
19
comment Sources on a category of ordinals
Ah, so what the author meant is a system whose vertices are sets of integers, rather than an integer-indexed system...
Jan
19
answered About stalk of sheaves in an unbalanced category
Jan
19
comment why algebraic structures?
No. A homomorphism need not be injective, and an injective function preserving constants, functions and relations need not be an embedding. You need to reflect relations as well.
Jan
19
comment What is an additive group?
Additive groups are typically assumed abelian.
Jan
19
comment Are concepts and properties studied in a category all preserved by morphisms?
(2) is a very vague question. What is studied depends on the category.
Jan
19
comment Are concepts and properties studied in a category all preserved by morphisms?
That's a rather weaker form of "preserving" than, say, quotient maps, where the preimage of a set is open if and only if the set itself is open.
Jan
19
comment Homotopy equivalence
(One should show that the free group on $m$ generators is not isomorphic to the free group on $n$ generators if $m \ne n$...)
Jan
19
comment Are concepts and properties studied in a category all preserved by morphisms?
No, obviously not, nor is it desirable that morphisms preserve "all" properties. For example a continuous map need not map open sets to open sets.
Jan
19
comment why algebraic structures?
A set is obviously an algebraic structure: in fact, its signature is trivial. One reason for restricting attention to signatures without relation symbols is that there are fewer subtleties about them. For instance, every injective homomorphism between two algebraic structures of the same signature is an embedding, but this is not true if you have relation symbols.
Jan
19
comment Composition of a kernel with a cokernel has a normal image
Hm, no, that doesn't sound right. Anyway $PNBE$ is not a pullback square either, but $P \to N$ is an isomorphism because $EBDC$ is a pullback square. (Use the pullback pasting lemma and pullback cancellation lemma.)