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8h
revised What are local homomorphisms, geometrically?
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8h
revised What are local homomorphisms, geometrically?
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9h
revised What are local homomorphisms, geometrically?
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9h
comment What are local homomorphisms, geometrically?
@zcn So, it appears to me that your proposition says the following: $A \to A / I$ reflects units if and only if the closed subset $V (I) \subseteq \operatorname{Spec} A$ contains all the closed points (of $\operatorname{Spec} A$). In particular, if $A \to B$ reflects units, then the scheme-theoretic image of $\operatorname{Spec} B \to \operatorname{Spec} A$ contains all closed points.
11h
comment What are local homomorphisms, geometrically?
However, do you have an explicit counterexample where there is a non-principal open subscheme of $Y$ through which $X \to Y$ factors? I was hoping that $X \to Y$ would be left orthogonal to arbitrary open immersions.
12h
comment What are local homomorphisms, geometrically?
Well spotted! I thought that it was some kind of surjectivity condition, but I couldn't quite figure out how.
13h
answered How do we get a simplicial homology functor?
14h
revised How do we get a simplicial homology functor?
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15h
comment How do we get a simplicial homology functor?
Replacing $\mathbf{Ab}$ with its skeleton only solves the issue of objects being defined up to isomorphism. You still have to worry about functoriality.
15h
asked What are local homomorphisms, geometrically?
1d
answered When are two morphisms of sheaves the same?
1d
answered Asterisk Notation
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comment Asterisk Notation
To be honest, I was half-expecting a follow-up question of, "what is the direct image functor?"
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comment Asterisk Notation
It is the direct image functor.
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comment Is the Axiom of Choice necessary to prove $\mathbb R \approx \mathcal P(\omega)$?
The order you define is called the lexicographic order. It is not a well-ordering. (Try finding an infinite descending chain!)
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comment Why are duals in a rigid/autonomous category unique up to unique isomorphism?
Yes. The same applies to limits, colimits, adjoints, etc.
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comment Why are duals in a rigid/autonomous category unique up to unique isomorphism?
That's not the point. A dual for $X$ is a triple $(Y, \epsilon, \eta)$ such that etc.; and for any two duals, there is a unique isomorphism connecting them that is compatible with all the data. That in no way says anything about the automorphisms of $Y$ as a bare object.
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comment Why are duals in a rigid/autonomous category unique up to unique isomorphism?
When we say "unique up to unique isomorphism", it must be understood in the right sense: namely, that there is a unique isomorphism of structured objects. For a much easier example, see here.
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comment The cone minus its apex deformation retracts onto its basis
Isn't $C (X) \setminus \{ P \}$ just homeomorphic to $X \times [0, 1)$?
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awarded  Nice Question