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| visits | member for | 2 years, 4 months |
| seen | 5 hours ago | |
| stats | profile views | 4,537 |
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Aug 15 |
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Understanding differential form @Tim: The (co)tangent bundle of a manifold is not automatically equipped with the structure of a vector space. I think it is very important to understand precisely what a vector bundle is and isn't before trying to understand what higher differential forms are. |
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Aug 15 |
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Bijective hom sets @Keenan: Oops, it seems I have assumed something extra. I was thinking about the fact that $f : G \to H$ is an epimorphism if and only if $\textrm{Hom}(f, K) : \textrm{Hom}(H, K) \to \textrm{Hom}(G, K)$ is injective for every $K$. As for group objects in the category of groups, essentially, the Eckmann–Hilton argument shows that any such must be an abelian group. |
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Aug 15 |
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Bijective hom sets @Keenan: Yes, but that's cheating. :p I was wondering about the following: Suppose $G$ is a group object in a category $\mathbf{C}$; given an epimorphism $f : G \to H$, does it follow that $H$ is also a group object, in a way making $f$ an internal homomorphism? As is well-known, an abelian group is simply a group object in the category of groups... |
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Aug 15 |
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Bijective hom sets It's quite easy to show that the condition on the hom-set map implies $f$ is an epimorphism. But at the moment I'm not seeing a good abstract-nonsense proof that epimorphisms transport abelian group structure... |
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Aug 15 |
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Canonical Isomorphism Between $\Omega^2(\mathbb{R^3})$ and $\mathbb{R^3}$? @Qiaochu: Actually, it's the sort of thing I might have said before I had learned about bundles and sheaves properly. So I can sympathise with the OP, a little. |
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Aug 14 |
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How to show $(a^b)^c=a^{bc}$ for arbitrary cardinal numbers? This seems like a very long answer for a very simple idea! But I've never tried to teach this to anyone, so perhaps it is necessary to have this much detail. |
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Aug 14 |
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How to prove cardinality equality ($\mathfrak c^\mathfrak c=2^\mathfrak c$) Hint: recall that $\mathfrak{c} = 2^{\aleph_0}$, so $\mathfrak{c}^{\mathfrak{c}} = 2^{\aleph_0 2^{\aleph_0}}$. |
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Aug 14 |
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$\mathbb R = X^2$ as a Cartesian product I echo Asaf's comments. For example, for simple cardinality reasons, $\mathbb{R} \times \mathbb{R}$ is in bijection with $\mathbb{R}$... and if you want to think of $\mathbb{R}$ as a concrete set in some axiomatic set theory, you need to be more specific. |
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Aug 14 |
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Definition of manifold Yes, exactly: the extra structure on the codomain is pulled back to give additional local structure on the domain. |
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Aug 13 |
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Proof of cardinality inequality What are your definitions of cardinal arithmetic, and what have you tried? For instance, under some definitions, all this amounts to is showing that from any pair of injective maps $M_1 \to M_2$ and $K_1 \to K_2$, you can get an injective map $K_1 \times M_1 \to K_2 \times M_2$... |
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Aug 13 |
accepted | Tensor products of infinite-dimensional spaces and other objects |
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Aug 13 |
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Tensor products of infinite-dimensional spaces and other objects @Qiaochu: Yes, I'm aware. My comment was informal: I was observing that, in some sense, $\textrm{Hom}(A, -)$ preserves products because of the product is defined to make it so. But it is a little distressing that the tensor product is neither a pure limit nor a pure colimit in most categories. |
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Aug 13 |
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Tensor products of infinite-dimensional spaces and other objects Your last point is interesting: I think the properties I've been ascribing to the hom functor are in fact properties of the product functor! (Because, after all, $\mathbf{Set}$ is also a monoidal closed category.) |
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Aug 13 |
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Tensor products of infinite-dimensional spaces and other objects @darij: Actually, I really didn't think through 2 and 3 very much at all. In the finite-dimensional case, for dimension reasons, 2 only works when $A$ is the base field, and even then it says something really trivial. |
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Aug 13 |
answered | Understanding isomorphic equivalences of tensor product |
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Aug 13 |
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$f^{\ast}\mathcal{F}\otimes_{\mathcal{O}_X}f^{\ast}\mathcal{G}\cong f^{\ast}(\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{G})\quad$? @Soarer: Yes, the stalk of a point is precisely the inverse image sheaf under the inclusion map $i : \{ * \} \to X$. |
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Aug 13 |
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Tensor products of infinite-dimensional spaces and other objects @Theo, Pierre-Yves: Thanks. Seems like I have a lot to (un)learn. |
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Aug 13 |
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Tensor products of infinite-dimensional spaces and other objects added 708 characters in body; edited tags |
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Aug 13 |
asked | Tensor products of infinite-dimensional spaces and other objects |
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Aug 13 |
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$f^{\ast}\mathcal{F}\otimes_{\mathcal{O}_X}f^{\ast}\mathcal{G}\cong f^{\ast}(\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{G})\quad$? I'm not sure whether this helps, but $f^{-1}$ preserves all small colimits (because it is a left adjoint) and all finite limits. But I'm not sure whether the tensor product can be constructed using just those... |
