Zhen Lin
Reputation
51,595
99/100 score
 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!