Zhen Lin
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 Dec 20 answered Does There Exist an Induced Model Strucutre via Ordinary Equivalence? Dec 20 comment Pointed objects in a category @MartinBrandenburg It suffices to assume that the coproduct insertion $Y \to Y \amalg 1$ is monic, which happens in e.g. any extensive category. Dec 20 comment Pointed objects in a category Hovey's book is known to have mistakes. You can find errata here. Dec 20 comment Pointed objects in a category The question has nothing to do with model categories. Also, in case it wasn't clear, my rings have unit and homomorphisms preserve them. So the only possible ring homomorphisms $\{ 0 \} \to R$ are those where $R$ is also trivial. Dec 20 answered Pointed objects in a category Dec 20 revised Pointed objects in a category edited tags; edited title Dec 20 awarded Popular Question Dec 20 comment Top de Rham cohomology Well, every differential $n$-form is closed, and if $M$ is compact, we can integrate any differential $n$-form over all of $M$. Since $M$ is orientable, it has a volume form, say $\omega$. Then for any other differential $n$-form $\alpha$, we can consider $\beta = \alpha - (\frac{1}{V} \int_M \alpha) \omega$, where $V = \int_M \omega$. Clearly, $\int_M \beta = 0$. So then it suffices to show that differential $n$-forms with vanishing integral are exact. Dec 19 awarded Enlightened Dec 19 awarded Nice Answer Dec 18 comment Does Hartshorne *really* not define things like the composition or restriction of morphisms of schemes? There is surely only one sensible way of defining composition. What else could there be? Dec 17 revised Unprovable Equivalence in Type Theory added 1977 characters in body Dec 17 answered Unprovable Equivalence in Type Theory Dec 17 revised Unprovable Equivalence in Type Theory edited tags Dec 16 answered The “closed” subspaces of topological algebraic structures Dec 16 comment Hartshorne's t functor This is the soberification functor for topological spaces. Dec 15 comment Hartshorne's definition of structure sheaf Incidentally, coproducts do exist in $\mathbf{CRing}$, and finite coproducts are tensor products. Dec 15 comment Proof that Beck-Chevalley holds for right adjoints iff it holds for left adjoints Oops. Fixed, thanks. Dec 15 revised Proof that Beck-Chevalley holds for right adjoints iff it holds for left adjoints edited body Dec 15 answered Proof that Beck-Chevalley holds for right adjoints iff it holds for left adjoints