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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