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Apr
14
comment Finding a formula for a $C^{\infty}$ 1-form $\omega$.
Use the chain rule for differentials: $d x^j = \sum \frac{\partial x^j}{\partial y^i} d y^i$.
Apr
14
comment Topological construct
It's not that simple. The left adjoint has a universal property with respect to all $A$, here we only want a universal property with respect to the given $A$ (or rather, the given source).
Apr
12
comment A map $f: X\rightarrow Y$ is a homotopy equivalence if and only if $h\circ f,f\circ k$ are homotopy equivalences of $X,Y$ respectively.
It's probably easiest to introduce the homotopy category, so as to avoid having to keep track of individual homotopies.
Apr
12
comment Too many independent cubic polynomials in an ideal $I\subset \mathbb C[x,y,z]$
Linearly independent over $\mathbb{C}$, or some other ring?
Apr
11
comment Derivatives on Functors
First of all you should tell us what you expect a derivative to behave like!
Apr
11
answered If the functor on presheaf categories given by precomposition by F is ff, is F full? faithful?
Apr
10
comment Combinatorial definition of the homotopy groups of a quasi category?
The fundamental group of a the nerve of a monoid is, if I'm not mistaken, precisely the group obtained by freely adjoining inverses.
Apr
10
comment Combinatorial definition of the homotopy groups of a quasi category?
Why not refer to "weak Kan complexes" as quasicategories? In which case it becomes clear that defining the fundamental group will at least include the process of making a group out of a monoid.
Apr
9
comment What does 'real-valued' function mean in topology?
Yes, it means exactly that. What else could it mean?
Apr
9
comment Total space of line bundle $\mathcal{O}(1)$ same as blow up of plane?
I think you're missing a dualisation somewhere. The total space of the tautological line bundle $\mathscr{O} (-1)$ on $\mathbb{P}^1$ is the blowup of affine plane at the origin, and accordingly there are no global sections. But $\mathscr{O} (1)$ has plenty of global sections.
Apr
8
comment Does defining the closure of a set as the intersection of all closed set that contain it requires the axiom of choice?
Perhaps you're uncomfortable with impredicativity.
Apr
8
comment Can the image of chains on a smooth manifold be thought of as a Borel $\sigma$-algebra?
You can consider the image if you like. But then you are not talking about chains anymore.
Apr
8
comment Can the image of chains on a smooth manifold be thought of as a Borel $\sigma$-algebra?
That is not the correct definition. An $n$-chain in a manifold $M$ is a formal sum (with integer or real coefficients, according to taste) of finitely many smooth maps $\Delta^n \to M$.
Apr
8
comment Can the image of chains on a smooth manifold be thought of as a Borel $\sigma$-algebra?
No. Chains are things you integrate over, and for a start, they are not subsets of the manifold.
Apr
8
comment functors on Zero-Object in $_RMod$-category
Well, $F$ would preserve zero objects, but only up to isomorphism.
Apr
8
comment functors on Zero-Object in $_RMod$-category
No. For instance, take a functor that maps every object to $S$ and every morphism to $0$...
Apr
8
comment Exact functors in the category of left R-modules - “Fun for the whole family”
Proceed in several steps. First prove that, in either definition, $T 0 = 0$. Then show that $T$ preserves binary products and kernels (in both definitions). Finally, show that a functor that preserves $0$, binary products, and kernels is left exact (in both definitions).
Apr
8
comment Surjection Vs Surjective geometric morphism
The right definition of "surjective geometric morphism", in my mind, is the one that says $f^*$ is conservative. See Lemma 3 later in the same section.
Apr
8
comment Surjection Vs Surjective geometric morphism
What's wrong with their proof? It's very nice and simple.
Apr
8
comment Can the image of chains on a smooth manifold be thought of as a Borel $\sigma$-algebra?
There is a world of difference between measurable sets and chains! Just because some people choose to write the operations in a $\sigma$-algebra using $+$ and $-$ and so on doesn't mean it's an abelian group under those operations!