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Jan
17
comment What's the “real” reason a finite map has finite fibers?
Of course, here we are using the fact that the class of finite morphisms is closed under base change.
Jan
17
comment Is it possible to characterize the theory of Integral domains with first-order logic alone ?
It is standard to have the axiom $0 \ne 1$ in the definition of integral domains.
Jan
16
answered Extending a Homotopy from $X \times [0,1]$ to $X \times \mathbb{R}$
Jan
15
revised Demonstrate currying via homomorphism
edited tags
Jan
14
revised Constant presheaf is not a sheaf
edited tags
Jan
14
comment Fibration with CW-complex as basespace admits retraction
Yes, the square commutes, so there is a lift. It's really that simple!
Jan
14
comment Fibration with CW-complex as basespace admits retraction
It works perfectly. No?
Jan
13
comment Is the existence of finite biproducts a strengthening of commutativity?
It's not so hard to work out if you know some basic facts.
Jan
13
answered Is the existence of finite biproducts a strengthening of commutativity?
Jan
13
accepted What is the algebraic tangent cone really?
Jan
13
comment Properties of a modified Zariski topology
Well, if you have both $z$ and $\bar{z}$, then you can separate out the real and imaginary parts of $z$. So I would guess that this is the same as the Zariski topology on $\mathbb{R}^{2 n}$.
Jan
13
comment Equivalent definition of Schemes
It is an interesting idea, but it's not at all obvious that schemes in the usual sense have this property.
Jan
13
comment Equivalent definition of Schemes
This is not at all how schemes are defined. Only some colimits define schemes.
Jan
12
comment Frobenius map in scheme theory
There's no difference, because roots of unity are always invertible. The two kernels are isomorphic as $k$-schemes but have different group structures.
Jan
12
answered Are units in rigid (autonomous) categories some sort of natural transformation?
Jan
12
revised Frobenius map in scheme theory
edited tags
Jan
12
answered Frobenius map in scheme theory
Jan
12
comment Show that function $f: A \to B$ is surjective when there is an implication: $g \circ f = h\circ f \to g=h$
Essentially. Go back to the axiom of separation: we have $x \in h (b)$ if and only if $b \in B'$. Of course, $h (b) \subseteq 1 = \{ 0 \}$, so it comes down to $0 \in h (b)$ if and only if $b \in B'$. If $b \notin B'$ then $0 \notin h (b)$, so $h (b) \ne 1$.
Jan
12
comment Open coverings and (co)limits
Yes, see here.
Jan
12
comment Show that function $f: A \to B$ is surjective when there is an implication: $g \circ f = h\circ f \to g=h$
It's an application of the axiom of separation. The only funny thing is that $x$ does not appear on the RHS, but that's allowed.