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Feb
21
answered How to interpret $1 \to 0$ in ${\bf Set}^\mathrm{op}$, and ${\bf Set}^\mathrm{op}$ itself?
Feb
21
comment Examples of categories where morphisms are not functions
If you soften your "and" to "or" then that's easy: take the category of sets and relations. But it turns out to be isomorphic to a different category whose objects are structured sets and whose morphisms are functions. There is also a locally small category which is demonstrably not equivalent to any concrete category. See also here.
Feb
20
comment Example where Čech and derived functor cohomologies don't agree.
More seriously, one can prove without any hypotheses at all that Čech cohomology (after taking the direct limit over all open covers) always computes the correct $H^1$, so any counterexample has to be in $H^2$ or higher.
Feb
20
comment Example where Čech and derived functor cohomologies don't agree.
A scheme is in particular a topological space, and their sheaf cohomology only depends on the topology. In fact, they are very simple as topological spaces. :p
Feb
20
revised Example where Čech and derived functor cohomologies don't agree.
deleted 18 characters in body; edited tags; edited title
Feb
20
comment adjunction relation
My definition just a functorial way of saying, take a coproduct of such-and-such-many copies of so-and-so. (It is a left adjoint, by the way.)
Feb
20
comment What is the notation for set of prime factors of a number?
Essentially, $\operatorname{Spec} \mathbb{Z}/(n)$.
Feb
20
comment Isomorphism between (source of) kernels of parallel arrow of a pullback square, by adjunction
Actually, isn't $(X \times_S S') \times_{S'} T \cong X \times_S T$ the pullback pasting lemma? This isomorphism has a somewhat different flavour than, say, $\ker (f \times g) \cong \ker (f) \times \ker (g)$.
Feb
20
comment Why do we accept Kuratowski's definition of ordered pairs?
Well, if you make some assertions about truncating the universe at some small rank or something silly like that, it might be important what kind of pairing function you use... (There's a comment in Hodges's Model theory about someone using definitions which made Lemma 4.3.1 false.)
Feb
20
revised Why do we accept Kuratowski's definition of ordered pairs?
edited body
Feb
20
comment Why do we accept Kuratowski's definition of ordered pairs?
You should say what $s$ does to sets that are not finite ordinals.
Feb
20
answered Products and pullbacks imply equalizers?
Feb
19
comment Freyd's Geometric Finiteness : An Example Computation
No, continuous does not mean preserving arbitrary joins. It would be more precise to say that it preserves "jointly epimorphic families", in the sense that if $\mathfrak{U} = \{ U_i \to V \}$ is an open cover of $V$, then the image of $\mathfrak{U}$ under $F$ is jointly epimorphic. However it is true that a left exact functor $F : \mathcal{O}(\mathbb{R}) \to \mathcal{B}$ must have image in the subterminals of $\mathcal{B}$, so in this case it is just a matter of preserving joins.
Feb
18
comment Show that the ideal of $k[X_1, X_2, X_3]$ generated by $X_1^3-X_3$ and $X_2^2-X_3$is a prime ideal.
I am referring to the notion of a graded ring. But perhaps your supervisor has a better solution; I'm not a professional algebraic geometer.
Feb
18
comment Show that the ideal of $k[X_1, X_2, X_3]$ generated by $X_1^3-X_3$ and $X_2^2-X_3$is a prime ideal.
You can use a kind of degree argument where $X_1$ has degree $2$, $X_2$ has degree $3$, and $X_3$ has degree $6$. (Also, are you one of my students?)
Feb
18
comment An example of decomposing a projective variety
This one is somewhat difficult. You have probably seen the solution by now, but if not, I'll just quickly say that one component is the line $V(X_2, X_3)$, and the other is the twisted cubic. (Also, the variety is supposed to be considered in $\mathbb{P}^3$.)
Feb
18
answered Show that the ideal of $k[X_1, X_2, X_3]$ generated by $X_1^3-X_3$ and $X_2^2-X_3$is a prime ideal.
Feb
18
comment To what extent is a scheme morphism determined by its topological map?
In the same vein, it is not quite true that a scheme morphism $X \to Y$ is determined by its action on points even if $X$ and $Y$ are varieties over some algebraically closed field $k$. You also have to know that the morphism commutes with the structural morphisms down to $\operatorname{Spec} k$.
Feb
18
comment Why isn't this a valid argument to the “proof” of the Axiom of Countable Choice?
No, there is certainly no problem with creating finite choice functions. Everyone has explained this. The problem is in assembling them together to make an infinite choice function.
Feb
18
comment Why isn't this a valid argument to the “proof” of the Axiom of Countable Choice?
@AloizioMacedo More to the point, mathematics is not temporal. A variable is not undefined at one time and then defined at another, so that you can jump over infinitely many steps and have them all simultaneously defined. Even if you could perform infinitely many tasks and jump ahead to a time when they have all been completed, that does not mean that things will be in a consistent state at that point. For example, consider the process where I turn on a lamp at $t = 0$, turn it off at $t = \frac{1}{2}$, turn it on again at $t = \frac{3}{4}$, etc.; is the lamp on or off at $t = 1$?