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8h
comment Correct definition of model category
@QiaochuYuan I just looked it up. Hovey apparently distinguishes between a category with a Model structure and a Model category. The latter is a category having all small limits/colimits, equipped with a Model structure.
8h
revised In a model category, is the full subcategory of fibrant objects a reflective subcategory?
added 23 characters in body
8h
comment In a model category, is the full subcategory of fibrant objects a reflective subcategory?
@DanielGerigk On the other hand, according to Hovey, model categories do come equipped with strong factorization systems. Hence, I believe the validity of the above statement is a matter of semantics.
9h
awarded  Student
9h
comment Correct definition of model category
@Kevin Carlson Yes, I think he says that in all the examples he considers, the factorization can be made functorial. I also know he uses the fact frequently. I don't think a weak factorization system on a model category always implies the existence of a functorial factorization. I'm looking for examples where statements cannot be proved without using functoriality.
10h
asked Correct definition of model category
2d
answered In a model category, is the full subcategory of fibrant objects a reflective subcategory?
Jan
15
revised Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$
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Jan
15
comment Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$
ok I've edited the post.
Jan
15
revised Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$
added 737 characters in body
Jan
15
comment Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$
@Kika Maybe I'm wrong, but I thought you can just calculate the eigenvalues of the adjoint. I think the conjugates give you the rest of the spectrum.
Jan
15
revised Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$
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Jan
15
answered Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$
Jan
15
comment Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$
What exactly is the operator? If your summing over $j$'s, $T((a_j))$ is not a sequence.
Dec
15
awarded  Necromancer
Dec
8
revised Can curvature be defined in Topos Theory?
edited body
Dec
6
awarded  Revival
Dec
6
revised Can curvature be defined in Topos Theory?
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Dec
6
answered Can curvature be defined in Topos Theory?
Dec
5
awarded  Yearling